work,energy and power - kumar's maths revision | p a g e work work done by a constant force if...
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Mechanics 2
Work, Energy and Power
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Work, Energy and Power
WORK ................................................................................................................................ 3
WORK DONE AGAINST GRAVITY ........................................................................... 4
WORK DONE AGAINST FRICTION ......................................................................... 5
QUESTIONS 1 ................................................................................................................ 7
FORCES AT AN ANGLE TO THE DIRECTION OF MOTION ............................ 9
ENERGY ........................................................................................................................... 10
KINETIC ENERGY ........................................................................................................ 10
POTENTIAL ENERGY .................................................................................................. 13
QUESTIONS 2 ............................................................................................................. 15
CONSERVATION OF ENERGY ................................................................................. 16
EXAM TYPE QUESTION .......................................................................................... 20
POWER ........................................................................................................................... 22
QUESTIONS 3 ............................................................................................................ 25
PAST EXAMINATION QUESTION ...................................................................... 27
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Work
Work done by a constant force
If the point of application of a force of F Newtons moves through a
distance s meters in the direction of the force then the work done by the
force is given by:
Work = F x s
The unit of work is the joule (J)
Example 1
The figure shows a box which is pulled at a constant speed across a
horizontal surface by a horizontal rope. When the box has moved a distance
of 9m the work done is 54 J. Find the constant resistance to the motion.
Work = force × distance moved
54 = force × 9
Resistance to motion = 6N
9m
R R
S
F F
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Work done against gravity
To raise a particle of mass m kg vertically at a constant speed you need to
apply a force of mg Newtons vertically upwards. If the particle is raised a
distance of h meters the work done against gravity is mgh joules. Work is
done against gravity when the particle is moving either vertically or at an
angle to the horizontal, for example on an inclined plane, but not when the
particle is moving along a horizontal plane.
Example 2
Find the work done by a child of mass 16kg whilst climbing to the top of a
slide of vertical height 9m.
Work done against gravity = mgh
= 16 × 9.8 × 9
= 1410J
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Work Done Against Friction
Example 3
A horizontal force P pulls a body of mass 2.75kg a distance of 12m across a
rough horizontal surface, coefficient of friction 0.2. The body moves with a
constant velocity and the only resisting force is that due to friction. Find
the work done against friction.
Resolving perpendicular to the plane:
R = 2.75g
Using F = μR
F = 0.25 × 9.8 × 2.75
F = 6.74N
Therefore:
Work = force × distance moved
= 6.74 × 12
= 80.9J
F
R
12m
P P
2.75g
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Exam questions will involve work done against gravity and friction.
Example 4
A particle of mass 8 kg is pulled at constant speed a distance of 28m up a
rough plane which is inclined at 30° to the horizontal. The coefficient of
friction between the particle and the surface is 0.2. Assuming the particle
moves up a line of greatest slope, find:
(a) the work done against friction.
(b) the work done against gravity.
(a) Work done against friction = force × distance moved
Resolving perpendicular to the plane gives:
R = 8g cos30º
Using F = μR F = 0.2 × 8g cos30º
F = 13.58N
Work done against friction = 13.58 × 28
= 380J
(b) Work done against gravity = mgh
Remember that we need to find the vertical displacement
Work done against gravity = 8 × g × 28 sin30º
=1097.6
=1100J
8g
F
P
30º
R
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Questions 1
1 A man building a wall lifts 75 bricks through a vertical distance of
3.5m. If each brick weighs 5kg, how much work does the man do
against gravity?
2 Find the work done against gravity when a person of mass 90kg climbs
a vertical distance of 32m.
3 A box of mass 12kg is pulled a distance of 25m across a horizontal
surface against resistances totaling 50N. If the body moves with
uniform velocity, find the work done against the resistances.
4 A horizontal force pulls a body of mass 6kg a distance of 11m across a
rough horizontal surface, coefficient of friction 0.25. The body moves
with uniform velocity and the only resisting force is that due to
friction. Find the work done.
5 A horizontal force pulls a body of mass 2.5kg a distance of 25m
across a rough horizontal surface, coefficient of friction 0.3. The
body moves with uniform velocity and the only resisting force is that
due to friction. Find the work done.
6 A smooth surface is inclined at an angle of 30º to the horizontal. A
parcel of mass 18kg lies on the surface and is pulled at a uniform
speed a distance of 7.5m up a line of greatest slope. Find the work
done against gravity.
7 A surface is inclined at an angle 1 3sin
5
to the horizontal. A body of
mass 78kg lies on the surface and is pulled at a uniform speed a
distance of 7.5m up a line of greatest slope against resistances
totaling 60N. Find:
a) the work done against gravity.
b) the work done against the resistances.
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8 A rough surface is inclined at an angle 1 12cos
13
to the horizontal. A
body of mass 140kg lies on the surface and is pulled at a uniform
speed a distance of 45m up the surface by a force acting along the
line of greatest slope. The coefficient of friction between the body
and the surface is 2
7
Find:
a) the frictional force acting.
b) the work done against friction.
c) the work done against gravity.
9 A rough surface is inclined at an angle 1 3
cos2
to the horizontal. A
body of mass 90kg lies on the surface and is pulled at a uniform speed
a distance of 15m up the surface by a force acting along a line of
greatest slope. The coefficient of friction between the body and the
surface is 1
8
Find:
a) the work done against friction.
b) the work done against gravity.
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Forces at an angle to the direction of motion
Consider a particle P resting on a horizontal surface. If a force of magnitude
F inclined at an angle θ to the horizontal causes the particle P to move along
the surface while remaining in contact with the surface, you can resolve the
force into its horizontal and vertical components.
For a force at an angle to the direction of motion:
Work done = component of force in direction of motion x distance moved in
the same direction.
Example 5
A sledge is pulled across a smooth horizontal floor by a force of magnitude
50 N inclined at 35° to the horizontal. Find the work done by the force in
moving the packing case a distance of 23m.
The horizontal distance is 23m so we must consider the horizontal
component of the force.
Horizontal component = 50 × cos35º = 40.96N
Work done = force × distance moved
= 40.96 × 23 = 942J
35º
50N
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Energy
The energy of a body is a measure of the capacity which the body has to do
work. When a force does work on a body it changes the energy of the body.
Energy exists in a number of forms, but we will consider two main types:
kinetic energy and potential energy.
Kinetic energy
The kinetic energy of a body is the energy that it possesses by virtue of its
motion. When a force acts on a body to increase its speed, then the work
done equates to the increase in kinetic energy of the body (provided that no
other forces are involved).
If a constant force F acts on a body of mass m, which is initially at rest on a
smooth horizontal surface, then after a distance s the body has velocity v.
So by considering the work equation:
Work done against friction = force × distance moved
= F × s
Using the fact that F = ma
and from the constant acceleration equations
2 2
2
v u 2as
u 0
va
2s
Therefore: work done = m × 2v
2s × s
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= 2mv
2
Therefore 2mv
2is said to be the kinetic energy of a mass m moving with
velocity v.
Example 6
A particle of mass 0.25kg is moving with a speed of 17ms . Find its kinetic
energy.
Using the formula:
21KE mv
2:
21KE 0.25 7
2
KE 6.13J
Example 7
A particle of mass 4 kg is being pulled across a smooth horizontal surface by
a horizontal force. The force does 46 J of work in increasing the particle's
velocity from 13ms to1pms. Find the value of p.
The change in Kinetic Energy is given by the formula:
2 21KE m(v u )
2
Where u is the initial and v is the final, velocity respectively. Since there
are no other forces involved the change in Kinetic Energy must equal the
work done by the force.
2 2
2 2
146 4(v 3 )
2
23 v 3
v 32 4 2
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Example 8
A van of mass 1600 kg starts from rest at a set of traffic lights. After
travelling 240 m its speed is 123ms . Given that the car is subject to a
constant resistance of 450 N find the constant driving force.
Initially the Kinetic Energy is zero.
Finally the Kinetic Energy is:
2
2
1KE mv
2
11600 23
2
423200J
Work done against the resistance = force × distance moved
= (F – 450) × 240
The change in Kinetic Energy equates to the work done. Therefore:
423200 (F 450) 240
F 1460N
F 450N
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Potential energy
If the particle is raised a distance of h meters the work done against
gravity is mgh joules. The work done against gravity equates to the increase
in potential energy. If the particle is lowered then the potential energy
decreases. When working with potential energy questions it is vital that a
zero potential energy point is decided upon.
Remember: P.E. = mgh
Example 9
A child of mass 14 kg is raised vertically through a distance of 1.8 m. Find
the increase in potential energy.
P.E. = mgh
= 14 × 9.8 × 1.8
= 247J
Example 10
A child of mass 40kg slides 5.5m down a playground slide inclined at an angle
of 4
arcsin7
to the horizontal. Model the child as a particle and the slide as
an inclined plane and hence calculate the potential energy lost by the child.
The important thing to remember is that potential energy changes after a
change in height. The 5.5m displacement is not the value of h.
5.5m
θ 40g
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4arcsin
7
means that sin θ = 4
7
.
h = 5.5 × sin θ
h = 5.5 × 4
7
= 3.142m
Therefore:
PE = mgh
= 40 × 9.8 × 3.142
= 1230J
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Questions 2
1 A body of mass 7.5kg, initially moving with velocity 4 1ms , increases
its kinetic energy by 55J. Find the final speed of the body.
2 Find the increase in kinetic energy when a stationary van of mass
1400kg accelerates at 4 2ms for 6 seconds.
3 Find the loss in kinetic energy when a car decelerates from 45 1kmh
to rest (take care with units).
4 Find the potential energy lost by a lift of mass 750kg as it descends
65m.
5 A body of mass 14kg, initially moving with a speed of 18 1ms ,
experiences a constant retarding force of 15N for 5 seconds. Find the
kinetic energy of the body at the end of this time.
6 A ski jumper of mass 95kg sets off from the top of the run and
travels a distance of 120. If the run is inclined at an angle of 42º to
the horizontal find the loss in potential energy.
120m
42º 95g
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Conservation of Energy
Consider the situation of a moving body where no work is done against
friction and that gravity is the only other force present then:
Total Energy = Kinetic Energy + Potential Energy = Constant
Or in other terms:
Total Initial Energy = Total Final Energy
Example 11
A particle of mass 3.5kg is released from rest and slides down a smooth
plane inclined at 4
arcsin7
to the horizontal. Find the distance travelled
while the particle increases its velocity to 5ms-1. Let the distance travelled
be z m.
Always consider PE and KE at the start and finish.
Initial KE = 0
2
2
1Final KE mv
2
13.5 5
2
43.75J
zm
θ 3.5g
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Let the final position be the zero potential energy point.
Initial PE = mgh
= 3.5 × 9.8 × zsin θ
= 34.3 × z × 4
7
By the Conservation of Energy Principle
Initial (KE + PE) = Final (KE + PE)
0 + 34.3 × z × 4
7
= 43.75 + 0
z = 2.23m
Example 12
A ball of mass 1.5kg is projected up a rough plane inclined at an angle of 30º
to the horizontal with a speed of 4 ms-1. Given that the coefficient of
friction between the particle and the plane is 0.2 find the distance the
particle moves up the plane before coming to rest.
4ms-1
0ms-1
30º 1.5g
1.5g
xm
R
F
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Let the initial position be the zero PE point.
2 21 1Initial KE mv 1.5 4
2 2
Initial KE 12J
Final KE = 0
Final PE = mgh
= 1.5 × 9.8 × xsin 30º
= 7.35xJ
The difference between the initial energy and the final energy is the work
done against friction. Therefore:
Initial (KE + PE) - Final (KE + PE) = Work Done (1)
To calculate the friction we first need to find the normal reaction force.
Resolving perpendicular to the plane gives:
R = 1.5g cos30º
R = 12.73N
Using F = μR F = 12.73 × 0.2 = 2.55N
Work done = force × distance moved
= 2.55 × x
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Substituting all of the values into equation (1) gives:
( 12 + 0 ) – ( 7.35x) = 2.55x
12 = 9.9x
x = 1.21m
The above type of question appears regularly on M2 papers and it is usual to
be told that the question must be solved by application of the energy
principle. Otherwise constant acceleration equations could be used.
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Exam Type Question
Michael Johnson reaches the top of a hill with a speed of 16.5ms . He
descends 50 m and then ascends 28 m to the top of the next incline. His
speed is now 14.5ms . Michael has a mass of 83 kg. The total distance that
Michael runs is 400m, and there is a constant resistance to motion of 9 N.
By consideration of the energy principle find the work done by Michael.
Start by considering the KE and PE at the beginning of the race. Assume
that the zero PE point is at the end. Therefore we only need to consider a
22m vertical displacement when working out the change in the PE.
2 21 1Initial KE mv 83 6.5
2 2
Initial KE 1753J
Initial PE = mgh
= 83 × 9.8 × 22
= 17895J
Total Energy initially = 19648J
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2 21 1Final KE mv 83 4.5
2 2
Final KE 840J
Final PE = 0
Total Energy Finally = 840J
Therefore the change in energy = 18808J
This change in energy has obviously been brought about by work. The work
done against the resistance is:
Work done against resistance = 9 × 400 = 3600J
The difference between the change in energy and the work done against
resistances must be the work done by Michael
Work done by Michael = 18808 – 3600
= 15208J
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Power
Power is a measure of the rate at which work is being done. The unit of work
is the Watt. If 1 Joule of work is done in 1 second then the rate of work is 1
Watt.
Power = Force × Distance moved in one second
or
Power = Force × Velocity
Example 13
A force of magnitude 750 N pulls a car up a slope at a constant speed of 19ms . Given that the force acts parallel to the direction of motion find, in
kW, the power developed.
Power = Force × Velocity
= 750 × 9
= 6750 = 6.75kW
Example 14
A car of mass 1500kg is travelling along a level road against a constant
resistance of magnitude 425 N. The engine of the car is working at 6 kW.
Calculate:
(a) the acceleration when the car is travelling at 13ms
(b) the maximum speed of the car.
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(a) Using the power equation to find the engine force
Power = Force × Velocity
6000 = Force × 13ms
Force = 2000N
Applying the engine force to the diagram:
Using F = ma
2000 – 425 = 1500a
a = 1.05 2ms
(b) The maximum speed occurs when the acceleration is zero. At this point
the engine force equates to the resistance. By using the power equation
again:
Power = Force × Velocity
6000 = 425 × v
v = 114.1ms
Example 15
A car of mass 1400kg is moving up a hill of slope 1
arcsin12
at a constant
speed of 125ms . If the power developed by the engine is 32 kW find the
resistance to motion.
At the top of the hill the road becomes horizontal. Find the initial
acceleration, assuming the resistance to be unchanged.
13ms 425N
1500kg
2000N
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If the engine is working at a rate of 32kW the engine force FE can be found
Power = Force × Velocity
32000 = FE × 25
FE = 1280N
Since the velocity is constant the weight component of the car acting
downhill and the resistance to motion must equate to the engine force.
Therefore:
1400g sinα + R = 1280
sinα = 1
12
1143.33 + R = 1280
R = 137N
As the vehicle reaches the top of the hill only R is acting to slow the car
down. With the engine force still being 1280N we can use F = ma.
1280 – 137 = 1400 × a
a = 0.816ms-2
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Questions 3
1 An athlete is running on a level track at a constant speed of 9ms-1. If
the resistances to motion equal 45N find the rate at which the
athlete is working.
2 A motorcycle is driven along a horizontal road against a constant
resistance of 350N. Find the maximum speed achieved by the
motorcycle if the engine works at a rate of
a) 3kW b) 5kW c) 9kW
3 A van traveling at maximum speed travels along a horizontal road. If
the engine is working at a rate of 16kW and the speed attained is
42ms-1 calculate the magnitude of the resistances.
4 A cyclist and bike have a combined mass of 85kg. If the cyclist is
working at a constant rate of 250W against resistances totaling 18N
calculate the maximum speed.
The cyclist then ascends a slope of incline 10º. If the cyclist continues
to work at the same rate and the resistances remain unchanged
calculate the maximum speed up the incline.
5 A small van travels along a level road against constant resistance to
motion of 450N. The mass of the van is 1200kg and its maximum
speed is 35ms-1. Calculate the maximum speed of the same van on an
incline of 1
arcsin35
assuming that the resistance and the rate at
which the engine works remain unchanged.
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6 Bob is busy removing roof tiles from a house. He sends the tiles down
a chute of length 12m, inclined at an angle of 35º to the horizontal.
The roof tiles weigh 4.5kg each and are released from rest at the top
of the chute. When the tile reaches the bottom it has a speed of
6.5ms-1. Calculate:
a) the potential energy lost by the tile.
b) the frictional force, assumed constant, which acts on the tile.
c) the coefficient of friction between the tile and the chute.
If Wendy pushes a tile from the top of the chute with a speed of
1.5ms-1 what is its speed at the bottom?
7 Thomas the engine has a mass of 7.5 × 105kg and whilst traveling along
a horizontal track he experiences resistance to motion of magnitude
1.75 × 105N.
a) Calculate the force that Thomas produces when the acceleration is
1.5ms-2.
b) Calculate the power produced at the instant that Thomas is
traveling at 12ms-1.
c) If Thomas then works at a constant rate of 600kW, find his
greatest possible speed along the track.
8 A Milk float of mass 1500kg moves with a constant speed of 7.5ms-1
up a slope with incline 1
arcsin8
. Given that the engine is working at a
rate of 21kW, find the resistance to motion.
9 With its engine working at a constant rate of 10kW, a car of mass
750kg can descend a slope of incline 1 in 50 at twice the steady speed
that it can ascend the same slope with the resistance to motion
remaining constant. Find the resistance to motion and the speed of
ascent.
10 A car of mass 700kg is moving along a straight horizontal road against
a constant resistive force of magnitude 450N. The engine of the car
is working at a rate of 8.76kW.
a) find the acceleration of the car at the instant when its speed is
12ms-1.
The car now moves up a straight road inclined at an angle of
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1arcsin
20
to the horizontal against the same resistive force of
magnitude 450N. If the car moves at a constant speed of Vms-1 whilst
working at a rate of 12kW, find the value of V.
11 A lorry of mass 1500 kg moves along a straight horizontal road. The
resistance to the motion of the lorry has magnitude 750 N and the
lorry’s engine is working at a rate of 36 kW.
(a) Find the acceleration of the lorry when its speed is 20 ms1.
The lorry comes to a hill inclined at an angle to the horizontal,
where sin = 101 . The magnitude of the resistance to motion from non-
gravitational forces remains 750 N.
The lorry moves up the hill at a constant speed of 20 ms1.
(b) Find the rate at which the lorry's engine is now working.
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Past Examination Questions
1. A child is playing with a small model of a fire-engine of mass 0.5 kg and a straight, rigid
plank. The plank is inclined at an angle to the horizontal. The fire-engine is projected
up the plank along a line of greatest slope. The non-gravitational resistance to the motion
of the fire-engine is constant and has magnitude R newtons.
When = 20 the fire-engine is projected with an initial speed of 5 m s1 and first comes
to rest after travelling 2 m.
(a) Find, to 3 significant figures, the value of R.
(7)
When = 40 the fire-engine is again projected with an initial speed of 5 m s1.
(b) Find how far the fire-engine travels before first coming to rest.
(3)
June 2001, Q5
2. A particle of mass 4 kg is moving in a straight horizontal line. There is a constant
resistive force of magnitude R newtons. The speed of the particle is reduced from 25 m
s1 to rest over a distance of 200 m.
Use the work-energy principle to calculate the value of R.
(4)
Jan 2002, Q1
3. A van of mass 1500 kg is driving up a straight road inclined at an angle to the
horizontal, where sin = 12
1. The resistance to motion due to non-gravitational forces is
modelled as a constant force of magnitude 1000 N.
Given that initially the speed of the van is 30 m s1 and that the van’s engine is working
at a rate of 60 kW,
(a) calculate the magnitude of the initial decleration of the van.
(4)
When travelling up the same hill, the rate of working of the van’s engine is increased to
80 kW. Using the same model for the resistance due to non-gravitational forces,
(b) calculate in m s1 the constant speed which can be sustained by the van at this rate of
working.
(4)
(c) Give one reason why the use of this model for resistance may mean that your answer
to part (b) is too high.
(1)
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Jan 2002, Q2
4. Figure 1
A 8 m s1
B 5 m s1
20 m
12 m
Figure 1 shows the path taken by a cyclist in travelling on a section of a road. When the
cyclist comes to the point A on the top of a hill, she is travelling at 8 m s1. She descends
a vertical distance of 20 m to the bottom of the hill. The road then rises to the point B
through a vertical distance of 12 m. When she reaches B, her speed is 5 m s1. The total
mass of the cyclist and the cycle is 80 kg and the total distance along the road from A to B
is 500 m. By modelling the resistance to the motion of the cyclist as of constant
magnitude 20 N,
(a) find the work done by the cyclist in moving from A to B.
(5)
At B the road is horizontal. Given that at B the cyclist is accelerating at 0.5 m s2,
(b) find the power generated by the cyclist at B.
(4)
June 2002, Q3
5. A car of mass 1000 kg is moving along a straight horizontal road with a constant
acceleration of f m s2. The resistance to motion is modelled as a constant force of
magnitude 1200 N. When the car is travelling at 12 m s1, the power generated by the
engine of the car is 24 kW.
(a) Calculate the value of f.
(4)
When the car is travelling at 14 m s1, the engine is switched off and the car comes to
rest, without braking, in a distance of d metres. Assuming the same model for resistance,
(b) use the work-energy principle to calculate the value of d.
(3)
(c) Give a reason why the model used for the resistance to motion may not be realistic.
(1)
Jan 2003, Q2
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6. A girl and her bicycle have a combined mass of 64 kg. She cycles up a straight stretch of
road which is inclined at an angle to the horizontal, where 1
14sin . She cycles at a
constant speed of 5 m s1. When she is cycling at this speed, the resistance to motion
from non-gravitational forces has magnitude 20 N.
(a) Find the rate at which the cyclist is working.
(4)
She now turns round and comes down the same road. Her initial speed is 15 m s , and the
resistance to motion is modelled as remaining constant with magnitude 20 N. She free-
wheels down the road for a distance of 80 m. Using this model,
(b) find the speed of the cyclist when she has travelled a distance of 80 m.
(5)
The cyclist again moves down the same road, but this time she pedals down the road. The
resistance is now modelled as having magnitude proportional to the speed of the cyclist.
Her initial speed is again 5 m s1 when the resistance to motion has magnitude 20 N.
(c) Find the magnitude of the resistance to motion when the speed of the cyclist is 18 m s .
(1)
The cyclist works at a constant rate of 200 W.
(d) Find the magnitude of her acceleration when her speed is 18 m s .
(4)
June 2003, Q6
7. A car of mass 400 kg is moving up a straight road inclined at an angle to the horizontal,
where sin = 141
. The resistance to motion of the car from non-gravitational forces is
modelled as a constant force of magnitude R newtons. When the car is moving at a
constant speed of 20 m s1, the power developed by the car’s engine is 10 kW.
Find the value of R.
(5)
Jan 2004, Q1
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8. Figure 1
B
3m
A
30
A particle P of mass 2 kg is projected from a point A up a line of greatest slope AB of a
fixed plane. The plane is inclined at an angle of 30 to the horizontal and AB = 3 m with
B above A, as shown in Fig. 1. The speed of P at A is 10 m s1.
Assuming the plane is smooth,
(a) find the speed of P at B.
(4)
The plane is now assumed to be rough. At A the speed of P is 10 m s1 and at B the speed
of P is 7 m s1. By using the work-energy principle, or otherwise,
(b) find the coefficient of friction between P and the plane.
(5)
Jan 2004, Q3
9. A lorry of mass 1500 kg moves along a straight horizontal road. The resistance to the
motion of the lorry has magnitude 750 N and the lorry’s engine is working at a rate of 36
kW.
(a) Find the acceleration of the lorry when its speed is 20 m s1.
(4)
The lorry comes to a hill inclined at an angle to the horizontal, where sin = 101 . The
magnitude of the resistance to motion from non-gravitational forces remains 750 N.
The lorry moves up the hill at a constant speed of 20 m s1.
(b) Find the rate at which the lorry's engine is now working.
(3)
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June 2004, Q1
10.
In a ski-jump competition, a skier of mass 80 kg moves from rest at a point A on a ski-
slope. The skier’s path is an arc AB. The starting point A of the slope is 32.5 m above
horizontal ground. The end B of the slope is 8.1 m above the ground. When the skier
reaches B, she is travelling at
20 m s1, and moving upwards at an angle to the horizontal, where tan = 43 , as shown
in Fig. 2. The distance along the slope from A to B is 60 m. The resistance to motion
while she is on the slope is modelled as a force of constant magnitude R newtons. By
using the work-energy principle,
(a) find the value of R.
(5)
On reaching B, the skier then moves through the air and reaches the ground at the point
C. The motion of the skier in moving from B to C is modelled as that of a particle moving
freely under gravity.
(b) Find the time for the skier to move from B to C.
(5)
(c) Find the horizontal distance from B to C.
(2)
(d) Find the speed of the skier immediately before she reaches C.
(5)
June 2004, Q7
32.5 m
20 m s1
8.1 m
B
A
C
33 | P a g e
11. Figure 3
A small package P is modelled as a particle of mass 0.6 kg. The package slides down a
rough plane from a point S to a point T, where ST = 12 m. The plane is inclined at an
angle of 30 to the horizontal and ST is a line of greatest slope of the plane, as shown in
Figure 3. The speed of P at S is 10 m s–1 and the speed of P at T is 9 m s–1. Calculate
(a) the total loss of energy of P in moving from S to T,
(4)
(b) the coefficient of friction between P and the plane.
(5)
Jan 2005, Q3
12. A car of mass 1000 kg is towing a trailer of mass 1500 kg along a straight horizontal
road. The tow-bar joining the car to the trailer is modelled as a light rod parallel to the
road. The total resistance to motion of the car is modelled as having constant magnitude
750 N. The total resistance to motion of the trailer is modelled as of magnitude R
newtons, where R is a constant. When the engine of the car is working at a rate of 50 kW,
the car and the trailer travel at a constant speed of 25 m s–1.
(a) Show that R = 1250.
(3)
When travelling at 25 m s–1 the driver of the car disengages the engine and applies the
brakes. The brakes provide a constant braking force of magnitude 1500 N to the car. The
resisting forces of magnitude 750 N and 1250 N are assumed to remain unchanged.
Calculate
(b) the deceleration of the car while braking,
(3)
(c) the thrust in the tow-bar while braking,
(2)
12 m
S P
T
30
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(d) the work done, in kJ, by the braking force in bringing the car and the trailer to rest.
(4)
(e) Suggest how the modelling assumption that the resistances to motion are constant
could be refined to be more realistic.
(1)
Jan 2005, Q5
13. A car of mass 1200 kg moves along a straight horizontal road. The resistance to motion
of the car from non-gravitational forces is of constant magnitude 600 N. The car moves
with constant speed and the engine of the car is working at a rate of 21 kW.
(a) Find the speed of the car.
(3)
The car moves up a hill inclined at an angle to the horizontal, where sin = 141
.
The car’s engine continues to work at 21 kW and the resistance to motion from non-
gravitational forces remains of magnitude 600 N.
(b) Find the constant speed at which the car can move up the hill.
(4)
June 2005, Q1
14. At a demolition site, bricks slide down a straight chute into a container. The chute is
rough and is inclined at an angle of 30 to the horizontal. The distance travelled down the
chute by each brick is 8 m. A brick of mass 3 kg is released from rest at the top of the
chute. When it reaches the bottom of the chute, its speed is 5 m s–1.
(a) Find the potential energy lost by the brick in moving down the chute.
(2)
(b) By using the work-energy principle, or otherwise, find the constant frictional force
acting on the brick as it moves down the chute.
(5)
(c) Hence find the coefficient of friction between the brick and the chute.
(3)
Another brick of mass 3 kg slides down the chute. This brick is given an initial speed of
2 m s–1 at the top of the chute.
(d) Find the speed of this brick when it reaches the bottom of the chute.
35 | P a g e
(5)
June 2005, Q7
15. A brick of mass 3 kg slides in a straight line on a horizontal floor. The brick is modelled
as a particle and the floor as a rough plane. The initial speed of the brick is 8 m s–1. The
brick is brought to rest after moving 12 m by the constant frictional force between the
brick and the floor.
(a) Calculate the kinetic energy lost by the brick in coming to rest, stating the units of
your answer.
(2)
(b) Calculate the coefficient of friction between the brick and the floor.
(4)
Jan 2006, Q1
16. A car of mass 1000 kg is moving along a straight horizontal road. The resistance to
motion is modelled as a constant force of magnitude R newtons. The engine of the car is
working at a rate of 12 kW. When the car is moving with speed 15 m s–1, the acceleration
of the car is 0.2 m s–2.
(a) Show that R = 600.
(4)
The car now moves with constant speed U m s–1 downhill on a straight road inclined at
to the horizontal, where sin = 401
. The engine of the car is now working at a rate of 7
kW. The resistance to motion from non-gravitational forces remains of magnitude R
newtons.
(c) Calculate the value of U.
(5)
Jan 2006, Q3
17. A car of mass 1200 kg moves along a straight horizontal road with a constant speed
of 24 m s–1. The resistance to motion of the car has magnitude 600 N.
(a) Find, in kW, the rate at which the engine of the car is working.
(2)
The car now moves up a hill inclined at to the horizontal, where sin = 281
. The
resistance to motion of the car from non-gravitational forces remains of magnitude 600
N. The engine of the car now works at a rate of 30 kW.
(b) Find the acceleration of the car when its speed is 20 m s–1.
(4)
June 2006, Q2
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18. A particle P has mass 4 kg. It is projected from a point A up a line of greatest slope of a
rough plane inclined at an angle to the horizontal, where tan = 43. The coefficient of
friction between P and the plane is 72. The particle comes to rest instantaneously at the
point B on the plane, where AB = 2.5 m. It then moves back down the plane to A.
(a) Find the work done by friction as P moves from A to B.
(4)
(b) Using the work-energy principle, find the speed with which P is projected from A.
(4)
(c) Find the speed of P when it returns to A.
(4)
June 2006, Q7
19. A particle of mass 0.8 kg is moving in a straight line on a rough horizontal plane. The
speed of the particle is reduced from 15 m s–1 to 10 m s–1 as the particle moves 20 m.
Assuming the only resistance to motion is the friction between the particle and the plane,
find
(a) the work done by friction in reducing the speed of the particle from 15 m s–1 to 10 m
s–1,
(2)
(b) the coefficient of friction between the particle and the plane.
(4)
Jan 2007, Q1
20. A car of mass 800 kg is moving at a constant speed of 15 m s–1 down a straight road
inclined at an angle to the horizontal, where sin = 241
. The resistance to motion from
non-gravitational forces is modelled as a constant force of magnitude 900 N.
(a) Find, in kW, the rate of working of the engine of the car.
(4)
When the car is travelling down the road at 15 m s–1, the engine is switched off. The car
comes to rest in time T seconds after the engine is switched off. The resistance to motion
from non-gravitational forces is again modelled as a constant force of magnitude 900 N.
(b) Find the value of T.
(4)
Jan 2007, Q2
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21. A cyclist and his bicycle have a combined mass of 90 kg. He rides on a straight road up a
hill inclined at an angle to the horizontal, where sin = 211
. He works at a constant rate
of 444 W and cycles up the hill at a constant speed of 6 m s–1.
Find the magnitude of the resistance to motion from non-gravitational forces as he cycles
up the hill.
(4)
June 2007, Q1
22. Figure 2
Two particles A and B, of mass m and 2m respectively, are attached to the ends of a light
inextensible string. The particle A lies on a rough plane inclined at an angle to the
horizontal, where tan = 43. The string passes over a small light smooth pulley P fixed at
the top of the plane. The particle B hangs freely below P, as shown in Figure 2. The
particles are released from rest with the string taut and the section of the string from A to
P parallel to a line of greatest slope of the plane. The coefficient of friction between A
and the plane is 85. When each particle has moved a distance h, B has not reached the
ground and A has not reached P.
(a) Find an expression for the potential energy lost by the system when each particle has
moved a distance h.
(2)
When each particle has moved a distance h, they are moving with speed v. Using the
work-energy principle,
(b) find an expression for v2, giving your answer in the form kgh, where k is a number.
(5)
June 2007, Q4
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23. A parcel of mass 2.5 kg is moving in a straight line on a smooth horizontal floor. Initially
the parcel is moving with speed 8 m s–1. The parcel is brought to rest in a distance of 20
m by a constant horizontal force of magnitude R newtons. Modelling the parcel as a
particle, find
(a) the kinetic energy lost by the parcel in coming to rest,
(2)
(b) the value of R.
(3)
Jan 2008, Q1
24. A car of mass 1000 kg is moving at a constant speed of 16 m s–1 up a straight road
inclined at an angle θ to the horizontal. The rate of working of the engine of the car is 20
kW and the resistance to motion from non-gravitational forces is modelled as a constant
force of magnitude 550 N.
(a) Show that sin θ = 141
.
(5)
When the car is travelling up the road at 16 m s–1, the engine is switched off. The car
comes to rest, without braking, having moved a distance y metres from the point where
the engine was switched off. The resistance to motion from non-gravitational forces is
again modelled as a constant force of magnitude 550 N.
(b) Find the value of y.
(4)
Jan 2008, Q3
25. A lorry of mass 2000 kg is moving down a straight road inclined at angle to the
horizontal, where sin = 251
. The resistance to motion is modelled as a constant force of
magnitude 1600 N. The lorry is moving at a constant speed of 14 m s–1.
Find, in kW, the rate at which the lorry’s engine is working.
(6)
May 2008, Q1
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26.
Figure 1
A package of mass 3.5 kg is sliding down a ramp. The package is modelled as a particle
and the ramp as a rough plane inclined at an angle of 20° to the horizontal. The package
slides down a line of greatest slope of the plane from a point A to a point B, where AB =
14 m. At A the package has speed 12 m s–1 and at B the package has speed 8 m s–1, as
shown in Figure 1.
Find
(a) the total energy lost by the package in travelling from A to B,
(5)
(b) the coefficient of friction between the package and the ramp.
(5)
May 2008, Q3
27. A car of mass 1500 kg is moving up a straight road, which is inclined at an angle θ to the
horizontal, where sin θ = 141
. The resistance to the motion of the car from non-
gravitational forces is constant and is modelled as a single constant force of magnitude
650 N. The car’s engine is working at a rate of 30 kW.
Find the acceleration of the car at the instant when its speed is 15 m s–1.
(5)
Jan 2009, Q1
28. A block of mass 10 kg is pulled along a straight horizontal road by a constant horizontal
force of magnitude 70 N in the direction of the road. The block moves in a straight line
passing through two points A and B on the road, where AB = 50 m. The block is modelled
as a particle and the road is modelled as a rough plane. The coefficient of friction
between the block and the road is 74.
(a) Calculate the work done against friction in moving the block from A to B.
(4)
The block passes through A with a speed of 2 m s–1.
40 | P a g e
(b) Find the speed of the block at B. (4)
Jan 2009, Q3
29. A truck of mass of 300 kg moves along a straight horizontal road with a constant speed
of 10 m s–1. The resistance to motion of the truck has magnitude 120 N.
(a) Find the rate at which the engine of the truck is working.
(2)
On another occasion the truck moves at a constant speed up a hill inclined at θ to the
horizontal, where sin θ = 141
. The resistance to motion of the truck from non-gravitational
forces remains of magnitude 120 N. The rate at which the engine works is the same as in
part (a).
(b) Find the speed of the truck.
(4)
May 2009, Q3
30.
Figure 4
A particle P of mass 2 kg is projected up a rough plane with initial speed 14 m s–1, from a
point X on the plane, as shown in Figure 4. The particle moves up the plane along the line
of greatest slope through X and comes to instantaneous rest at the point Y. The plane is
inclined at an angle α to the horizontal, where tan α = 247
. The coefficient of friction
between the particle and the plane is 81.
(a) Use the work-energy principle to show that XY = 25 m.
(7)
After reaching Y, the particle P slides back down the plane.
(b) Find the speed of P as it passes through X.
(4)
May 2009, Q7
41 | P a g e
31. A particle of mass 0.5 kg is projected vertically upwards from ground level with a speed
of 20 m s–1. It comes to instantaneous rest at a height of 10 m above the ground. As the
particle moves it is subject to air resistance of constant magnitude R newtons. Using the
work-energy principle, or otherwise, find the value of R.
(6)
Jan 2010, Q3
32. A cyclist and her bicycle have a total mass of 70 kg. She cycles along a straight
horizontal road with constant speed 3.5 m s–1. She is working at a constant rate of 490 W.
(a) Find the magnitude of the resistance to motion.
(4)
The cyclist now cycles down a straight road which is inclined at an angle θ to the
horizontal, where sin θ = 141
, at a constant speed U m s–1. The magnitude of the non-
gravitational resistance to motion is modelled as 40U newtons. She is now working at a
constant rate of 24 W.
(b) Find the value of U.
(7)
Jan 2010, Q5
33. A particle P of mass 0.6 kg is released from rest and slides down a line of greatest slope
of a rough plane. The plane is inclined at 30° to the horizontal. When P has moved 12 m,
its speed is 4 m s–1. Given that friction is the only non-gravitational resistive force acting
on P, find
(a) the work done against friction as the speed of P increases from 0 m s–1 to 4 m s–1,
(4)
(b) the coefficient of friction between the particle and the plane.
(4)
June 2010, Q2
34. A car of mass 750 kg is moving up a straight road inclined at an angle θ to the horizontal,
where sin θ = 151
. The resistance to motion of the car from non-gravitational forces has
constant magnitude R newtons. The power developed by the car’s engine is 15 kW and
the car is moving at a constant speed of 20 m s–1.
(a) Show that R = 260.
(4)
42 | P a g e
The power developed by the car’s engine is now increased to 18 kW. The magnitude of
the resistance to motion from non-gravitational forces remains at 260 N. At the instant
when the car is moving up the road at 20 m s–1 the car’s acceleration is a m s–2.
(b) Find the value of a.
(4)
June 2010, Q4
35. A cyclist starts from rest and moves along a straight horizontal road. The combined mass
of the cyclist and his cycle is 120 kg. The resistance to motion is modelled as a constant
force of magnitude 32 N. The rate at which the cyclist works is 384 W. The cyclist
accelerates until he reaches a constant speed of v m s–1.
Find
(a) the value of v,
(3)
(b) the acceleration of the cyclist at the instant when the speed is 9 m s–1.
(3)
Jan 2011, Q1
36. A particle of mass 2 kg is moving with velocity (5i + j) m s–1 when it receives an impulse
of (–6i + 8j) N s. Find the kinetic energy of the particle immediately after receiving the
impulse.
(5)
Jan 2011, Q2
37. A car of mass 1000 kg moves with constant speed V m s−1 up a straight road inclined at
an angle θ to the horizontal, where sin θ = 301
. The engine of the car is working at a rate
of 12 kW. The resistance to motion from non-gravitational forces has magnitude 500 N.
Find the value of V.
(5)
June 2011, Q1
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38.
Figure 2
A particle P of mass 0.5 kg is projected from a point A up a line of greatest slope AB of a
fixed plane. The plane is inclined at 30° to the horizontal and AB = 2 m with B above A,
as shown in Figure 2. The particle P passes through B with speed 5 m s–1. The plane is
smooth from A to B.
(a) Find the speed of projection.
(4)
The particle P comes to instantaneous rest at the point C on the plane, where C is above B
and BC = 1.5 m. From B to C the plane is rough and the coefficient of friction between P
and the plane is .
By using the work-energy principle,
(b) find the value of .
(6)
June 2011, Q5
39. A cyclist and her cycle have a combined mass of 75 kg. The cyclist is cycling up a
straight road inclined at 5° to the horizontal. The resistance to the motion of the cyclist
from non-gravitational forces is modelled as a constant force of magnitude 20 N. At the
instant when the cyclist has a speed of 12 m s−1, she is decelerating at 0.2 m s−2.
(a) Find the rate at which the cyclist is working at this instant.
(5)
When the cyclist passes the point A her speed is 8 m s−1. At A she stops working but does
not apply the brakes. She comes to rest at the point B.
The resistance to motion from non-gravitational forces is again modelled as a constant
force of magnitude 20 N.
(b) Use the work-energy principle to find the distance AB.
44 | P a g e
(5)
Jan 2012, Q3
40. A car of mass 1200 kg pulls a trailer of mass 400 kg up a straight road which is inclined
to the horizontal at an angle , where sin = 141
. The trailer is attached to the car by a
light inextensible towbar which is parallel to the road. The car’s engine works at a
constant rate of 60 kW. The non-gravitational resistances to motion are constant and of
magnitude 1000 N on the car and 200 N on the trailer.
At a given instant, the car is moving at 10 m s–1. Find
(a) the acceleration of the car at this instant,
(5)
(b) the tension in the towbar at this instant.
(4)
The towbar breaks when the car is moving at 12 m s–1.
(c) Find, using the work-energy principle, the further distance that the trailer travels
before coming instantaneously to rest.
(5)
May 2012, Q6
41. A lorry of mass 1800 kg travels along a straight horizontal road. The lorry’s engine is
working at a constant rate of 30 kW. When the lorry’s speed is 20 m s−1, its acceleration
is 0.4 m s−2. The magnitude of the resistance to the motion of the lorry is R newtons.
(a) Find the value of R.
(4)
The lorry now travels up a straight road which is inclined at an angle α to the horizontal,
where sin α = 121
. The magnitude of the non-gravitational resistance to motion is R
newtons. The lorry travels at a constant speed of 20 m s−1.
(b) Find the new rate of working of the lorry’s engine.
(5)
Jan 2013, Q2
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42. The point A lies on a rough plane inclined at an angle θ to the horizontal, where sin θ =
25
24. A particle P is projected from A, up a line of greatest slope of the plane, with speed
U m s−1. The mass of P is 2 kg and the coefficient of friction between P and the plane is
12
5. The particle comes to instantaneous rest at the point B on the plane, where
AB =1.5 m. It then moves back down the plane to A.
(a) Find the work done against friction as P moves from A to B.
(4)
(b) Use the work-energy principle to find the value of U.
(4)
(c) Find the speed of P when it returns to A.
(3)
Jan 2013, Q5
43. A particle P of mass 3 kg moves from point A to point B up a line of greatest slope of a
fixed rough plane. The plane is inclined at 20° to the horizontal. The coefficient of
friction between P and the plane is 0.4.
Given that AB = 15 m and that the speed of P at A is 20 m s–1, find
(a) the work done against friction as P moves from A to B,
(3)
(b) the speed of P at B.
(4)
June 2013, Q2
44. A caravan of mass 600 kg is towed by a car of mass 900 kg along a straight horizontal
road. The towbar joining the car to the caravan is modelled as a light rod parallel to the
road. The total resistance to motion of the car is modelled as having magnitude 300 N.
The total resistance to motion of the caravan is modelled as having magnitude 150 N. At
a given instant the car and the caravan are moving with speed 20 m s–1 and acceleration
0.2 m s–2.
(a) Find the power being developed by the car’s engine at this instant.
(5)
(b) Find the tension in the towbar at this instant.
(2)
46 | P a g e
June 2013_R, Q1
45. A ball of mass 0.2 kg is projected vertically upwards from a point O with speed 20 m s–1.
The non-gravitational resistance acting on the ball is modelled as a force of constant
magnitude 1.24 N and the ball is modelled as a particle. Find, using the work-energy
principle, the speed of the ball when it first reaches the point which is 8 m vertically
above O.
(6)
June 2013_R, Q2
46.
Figure 3
A particle P of mass 2 kg is released from rest at a point A on a rough inclined plane and
slides down a line of greatest slope. The plane is inclined at 30° to the horizontal. The
point B is 5 m from A on the line of greatest slope through A, as shown in Figure 3.
(a) Find the potential energy lost by P as it moves from A to B.
(2)
The speed of P as it reaches B is 4 m s–1.
(b) (i) Use the work-energy principle to find the magnitude of the constant frictional
force acting on P as it moves from A to B.
(ii) Find the coefficient of friction between P and the plane.
(7)
The particle P is now placed at A and projected down the plane towards B with speed 3 m
s–1. Given that the frictional force remains constant,
(c) find the speed of P as it reaches B.
(4)
June 2014_R, Q5
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47. The points A and B are 10 m apart on a line of greatest slope of a fixed rough inclined
plane, with A above B. The plane is inclined at 25° to the horizontal. A particle P of mass
5 kg is released from rest at A and slides down the slope. As P passes B, it is moving with
speed
7 m s–1.
(a) Find, using the work-energy principle, the work done against friction as P moves
from
A to B.
(4)
(b) Find the coefficient of friction between the particle and the plane.
(5)
June 2014, Q8
48.
Figure 2
A particle P of mass 10 kg is projected from a point A up a line of greatest slope AB of a
fixed rough plane. The plane is inclined at angle to the horizontal, where tan = 12
5
and AB = 6.5 m, as shown in Figure 2. The coefficient of friction between P and the plane
is . The work done against friction as P moves from A to B is 245 J.
(a) Find the value of .
(5)
The particle is projected from A with speed 11.5 m s–1. By using the work-energy
principle,
(b) find the speed of the particle as it passes through B.
(4)
June 2015, Q8
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49. A car of mass 800 kg is moving on a straight road which is inclined at an angle θ to the
horizontal, where sin θ = 1
20. The resistance to the motion of the car from non-
gravitational forces is modelled as a constant force of magnitude R newtons. When the
car is moving up the road at a constant speed of 12.5 m s−1, the engine of the car is
working at a constant rate of 3P watts. When the car is moving down the road at a
constant speed of 12.5 m s−1, the engine of the car is working at a constant rate of P
watts.
(a) Find
(i) the value of P,
(ii) the value of R.
(6)
When the car is moving up the road at 12.5 m s−1 the engine is switched off and the car
comes to rest, without braking, in a distance d metres. The resistance to the motion of the
car from non-gravitational forces is still modelled as a constant force of magnitude R
newtons.
(b) Use the work-energy principle to find the value of d.
(4)
June 2016, Q2
50. A car has mass 550 kg. When the car travels along a straight horizontal road there is a
constant resistance to the motion of magnitude R newtons, the engine of the car is
working at a rate of P watts and the car maintains a constant speed of 30 m s–1. When the
car travels up a line of greatest slope of a hill which is inclined at θ to the horizontal,
where sin θ = 1
14, with the engine working at a rate of P watts, it maintains a constant
speed of 25 m s–1. The non-gravitational resistance to motion when the car travels up the
hill is a constant force of magnitude R newtons.
(a) (i) Find the value of R.
(ii) Find the value of P.
(8)
(b) Find the acceleration of the car when it travels along the straight horizontal road at
20 ms–1 with the engine working at 50 kW.
(4)
Jan 2014 IAL, Q3
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51. A trailer of mass 250 kg is towed by a car of mass 1000 kg. The car and the trailer are
travelling down a straight road inclined at an angle θ to the horizontal, where sin θ = 1
20
.
The resistance to motion of the car is modelled as a single force of magnitude 300 N
acting parallel to the road. The resistance to motion of the trailer is modelled as a single
force of magnitude 100 N acting parallel to the road. The towbar joining the car to the
trailer is modelled as a light rod which is parallel to the direction of motion. At a given
instant the car and the trailer are moving down the road with speed 25 m s–1 and
acceleration 0.2 m s–2.
(a) Find the power being developed by the car’s engine at this instant.
(6)
(b) Find the tension in the towbar at this instant.
(4)
June 2014 IAL, Q2
52. A car of mass 500 kg is moving at a constant speed of 20 m s–1 up a straight road inclined
at an angle 𝜃 to the horizontal, where sin 𝜃 = 1
20. The resistance to motion from
non-gravitational forces is modelled as a constant force of magnitude 150 N.
(a) Find the rate of working of the engine of the car.
(5)
When the car is travelling up the road at 20 m s–1, the engine is switched off. The car then
comes to instantaneous rest, without braking, having moved a distance d metres up the
road from the point where the engine was switched off. The resistance to motion from
non-gravitational forces is again modelled as a constant force of magnitude 150 N.
(b) Use the work-energy principle to find the value of d.
(4)
Jan 2015 IAL, Q2
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53.
A particle P of mass 6.5 kg is projected up a fixed rough plane with initial speed 6 m s–1
from a point X on the plane, as shown in Figure 1. The particle moves up the plane along
the line of greatest slope through X and comes to instantaneous rest at the point Y, where
XY = d metres. The plane is inclined at an angle 𝜃 to the horizontal, where tan 𝜃 = 5
12.
The coefficient of friction between P and the plane is 1
3.
(a) Use the work-energy principle to show that, to 2 significant figures, d = 2.7.
(7)
After coming to rest at Y, the particle P slides back down the plane.
(b) Find the speed of P as it passes through X.
(4)
June 2015 IAL, Q4
54. A car of mass 900 kg is travelling up a straight road inclined at an angle θ to the
horizontal, where sin θ = 1
25. The car is travelling at a constant speed of 14 m s–1 and the
resistance to motion from non-gravitational forces has a constant magnitude of 800 N.
The car takes
10 seconds to travel from A to B, where A and B are two points on the road.
(a) Find the work done by the engine of the car as the car travels from A to B.
(4)
When the car is at B and travelling at a speed of 14 m s–1 the rate of working of the
engine
of the car is suddenly increased to P kW, resulting in an initial acceleration of the car of
0.7 m s–2. The resistance to motion from non-gravitational forces still has a constant
magnitude of 800 N.
(b) Find the value of P.
(4)
Jan 2016 IAL, Q1
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55. A truck of mass 1800 kg is moving along a straight horizontal road. The engine of the
truck is working at a constant rate of 10 kW. The non-gravitational resistance to motion is
modelled as a constant force of magnitude R newtons. At the instant when the truck is
moving with speed 15 m s–1, the acceleration of the truck is 0.25 m s–2.
(a) Find the value of R.
(4)
The truck now moves up a straight road at a constant speed V m s–1. The road is inclined
at an angle θ to the horizontal, where sin θ = 1
14. The non-gravitational resistance to
motion is now modelled as a constant force of magnitude 30V newtons. The engine of the
truck is now working at a constant rate of 12 kW.
(b) Find the value of V.
(6)
June 2016 IAL, Q2
56.
Two particles P and Q, of mass 2 kg and 4 kg respectively, are connected by a light
inextensible string. Initially P is held at rest at the point A on a rough fixed plane inclined
at α to the horizontal ground, where sin α = 3
5. The string passes over a small smooth
pulley fixed at the top of the plane. The particle Q hangs freely below the pulley and 2.5
m above the ground, as shown in Figure 1. The part of the string from P to the pulley lies
along a line of greatest slope of the plane. The system is released from rest with the string
taut. At the instant when Q hits the ground, P is at the point B on the plane. The
coefficient of friction between
P and the plane is 1
4.
(a) Find the work done against friction as P moves from A to B.
(4)
(b) Find the total potential energy lost by the system as P moves from A to B.
(3)
52 | P a g e
(c) Find, using the work-energy principle, the speed of P as it passes through B.
(4)
Jan 2016 IAL, Q4
57.
Two particles P and Q, of mass 2m and 3m respectively, are connected by a light
inextensible string. Initially P is held at rest on a fixed rough plane inclined at θ to the
horizontal ground, where sin θ = 2
5. The string passes over a small smooth pulley fixed at
the top of the plane. The particle Q hangs freely below the pulley, as shown in Figure 1.
The part of the string from P to the pulley lies along a line of greatest slope of the plane.
At time t = 0 the system is released from rest with the string taut. When P moves the
friction between P and the plane is modelled as a constant force of magnitude 3
5mg. At
the instant when each particle has moved a distance d, they are both moving with speed v,
particle P has not reached the pulley and
Q has not reached the ground.
(a) Show that the total potential energy lost by the system when each particle has moved
a distance d is 11
5mgd.
(3)
(b) Use the work-energy principle to find v2 in terms of g and d.
(4)
When t = T seconds, d = 1.5 m.
(c) Find the value of T.
(2)
June 2016 IAL, Q3
53 | P a g e
58. A particle P of mass 4 kg is projected with speed 6 m s–1 up a line of greatest slope
of a fixed rough inclined plane. The plane is inclined at angle α to the horizontal,
where sin α = 1
7. The particle is projected from the point A on the plane and comes to
instantaneous rest at the point B on the plane, where AB = 10 m.
(a) Show that the work done against friction as P moves from A to B is 16 joules.
(4)
After coming to instantaneous rest at B, the particle slides back down the plane.
(b) Use the work-energy principle to find the speed of P at the instant it returns to A.
(3)
Oct 2016 IAL, Q3
59. A car of mass 1200 kg moves up a straight road. The road is inclined to the horizontal at
an angle α where sin α =
1
15. The car is moving up the road with constant speed 10 m s–1
and the engine of the car is working at a constant rate of 11 760 watts. The non-
gravitational
resistance to motion has a constant magnitude of R newtons.
(a) Find the value of R.
(4)
The rate of working of the car is now increased to 50 kW. At the instant when the speed
of the car is V m s–1, the magnitude of the non-gravitational resistance to the motion of
the
car is 700 N and the acceleration of the car is 1.5 m s–2.
(b) Find the value of V.
(6)
Jan 2017 IAL, Q1
60. A ball of mass 0.6 kg is projected vertically upwards with speed 22.4 m s–1 from a point
which is 1.5 m above horizontal ground. The ball moves freely under gravity until it
reaches the ground. The ground is soft and the ball sinks 2.5 cm into the ground before
coming to rest. The ball is modelled as a particle and the ground is assumed to exert
a constant resistive force of magnitude R newtons on the ball. Using the work-energy
principle, find, to 3 significant figures, the value of R.
(5)
Jan 2017 IAL, Q6