AS Mechanics

Unit 2 (PHYA2)

Newton’s Laws• Objects stay at rest or in uniform motion (velocity,

i.e., speed + direction) unless acted on by an unbalanced force.

Notes:• Add up the forces in each perpendicular direction (x, y, z) and

see if there is a resultant – i.e., non-zero force Fx, Fy or Fz.

• The Ancient Greeks believed bodies’ usual state was to be still. They didn’t understand friction. Newton & Galileo showed that moving bodies stay moving. Think satellites and airtracks…

NI

AR

W

constant v

Newton’s Laws• The force on a body is proportional to

its rate of change of momentum.

Notes:• We have defined the newton (N) so that ∑F = ∆p/∆t• Since p = mv, ∑F = ∆mv/∆t we have two cases

(a) m constant: ∑F = m∆v/∆t = ma(b) v constant: ∑F = v∆m/∆t

• Examples: space rocket, tractor pulling trailer, water through hole in reservoir wall, winch. Case (a) or (b)?

• This applies in each perpendicular direction (x, y, z), i.e., ∑Fx = max, ∑Fy = vy∆m/∆t, etc., etc.

NII

Newton’s Laws• “Objects stay at rest or move with uniform velocity

unless acted on by a force”– No forces– Balanced forces (i.e., resultant = 0)

• Bigger forces cause bigger accelerations, larger masses get smaller accelerations, hence F = ma– F is the RESULTANT force (F1 + F2 + ...)

– a is ALWAYS in the same direction as F (even if v isn’t)

• The rate of change of momentum of a body is proportional to the resultant force on it.– Hence F ~ mv/t– 1 N is defined so that it is ‘equal’ as well as ‘proportional’

NII

NI

Newton’s Laws• When body A exerts a force on body B, then

B exerts an equal (magnitude) and opposite (direction) a force on A.

Notes:

• FA–B = –FB–A

• 6 pairs to find here:• Four conditions to be a NIII pair

(a) same type of force,(b) same magnitude,(c) opposite directions,(d) acting on different objects.

NIII

Weight• Weight is a force• F = ma, W = mg

g:• the gravitational field strength, or• the acceleration due to gravity• is independent of m• is 9.81 Nkg–1 (or 9.81 ms–2) in London

Classic mechanics problems

• Lifting mass

Classic mechanics problems

• Man in a lift– Force on lift = T – mg = ma– If lift is stationary or travelling at

constant velocity, T = mg

Change in motion

accelerating deceleratingDirection of travel

Up T>mg T<mg

Down T<mg T<mg

Classic mechanics problems

• Masses over a frictionless pulley (Atwood’s machine)• Tension in rope T is same on both

sides

• m2g – T = m2a

• T – m1g = m1a

• (adding) m2g – m1g = m2a + m1a

• or

• Can think of this as both masses being accelerated by a force equal to the difference in their weights

T

T

gmm

mma

12

12

Classic mechanics problems

• Inclined slope, no friction• Force acting down slope is

mgsin• So a = gsin• Block slides down slope• Note that although it takes

longer to reach the bottom than if it had dropped vertically, its speed at the bottom is the same – can you explain why?

Classic mechanics problems

• Inclined slope, with friction (F)

• Force acting down slope is mgsin – F

• So mgsin – F = ma• This time the velocity at

the bottom of the slope will be less than the frictionless case.

• Try the questions on p.137

F

F

Terminal speed• Speed or velocity as

we’re only interested in the downwards direction

• Drag depends on the object’s shape and speed, and the viscosity of the fluid

• Maximum acceleration is when v = 0. Use the gradient of the v–t graph.

Horizontal motion

• Driving force < counter force: vehicle slows down

• Driving force = counter force: vehicle moves at constant velocity

• Driving force > counter force: vehicle speeds up

Counter force

Driving force Driving force – provided by rider/engine

Counter force – air resistance and friction

Stopping distances

• stopping = thinking + brakings = s1 + s2

s1 = vt0 (speed × reaction time)

s2 = v2/2a (from v2 = u2 + 2as) (or s2 = WD/Fbraking)

• Thinking distance v

• Braking distance v2

Impact time

• Impact time is the duration of an impact force

• Remember

• so impact time is

• acceleration is

2

vu

t

s

Average speed

vu

st

2

t

uva

Impact example

• A 500 kg car crashes into a lamp post. The car stops from a speed of 13 ms–1 in a distance of 1 m.

• t = ?

• a = ?

• F = ?

svu

st 15.0

13

22

gmst

uva 8.87.86

15.0

13 2

NmaF 433507.86500

Road safety

• Airbags inflate to slow the deceleration of the head and spread pressure over a wider area

• Seatbelts are designed to stretch, slowing the deceleration of the body

• Crumple zones in vehicles increase the deceleration time

• Padding inside crash helmet increases deceleration time

• Now try Qs on p. 145

Work• When you expend energy to exert a force which

moves an object, you are doing “work”.

• Work done = energy transferred

metres Newtons Joules

force theofdirection in the distance x Force donework

Is any work being done in these cases?

No work done

No work done

Work questions

• How much work is done when a mass of 3 kg is lifted vertically through 6 m?– Work = F × d = (3 × 10) × 6 = 180 J

• A hiker climbs a hill 300 m high. If she weighs 500 N calculate the work she does lifting her body to the top of the hill.– Work = F × d = 500 × 300

= 150,000 J or 150 kJ

Motion and force in different directions

• Work done = component of average force in direction of motion × distance moved in direction of force

• W = Fscos• If = 90°, W = 0.

Force–distance graphs

• Area under a force–distance graph is work done

Constant force

varying force

Force extending a spring

• The more you stretch it, the harder you have to pull

• Hooke’s Law:F = kx

• So work to extend to

L is area under graph

extension

forc

e

LFLW

2

1

F

Kinetic energy• KE is the result of work being done.

• Imagine a constant force F accelerating a mass m:

mF

t=0, u=0 v

m

t

v

t

uvavtt

vus

,

2

1

2

2

2

1

2mv

vt

t

mvFsW

t

mvmaF

Kinetic Energy

• Kinetic energy (KE) is the energy an object has due to its motion.

• So an object has more KE if:– it has a greater mass, or– it moves faster

• A lorry and a van both travel at 15 ms–1. The lorry has a mass of 2000 kg and the van a mass of 1000 kg. What KE does each have?

2

2

1mvKE

mass velocity

30 kJ

15 kJ

Kinetic Energy

• Calculate the KE of a 500 kg car travelling ata) 10 ms–1

b) 20 ms–1

• So what effect does doubling the speed have on the kinetic energy?

10 m/s

20 m/s25 kJ

100 kJ

It increases by a factor of 4

• How much work does the engine have to do to increase the speed from 10 to 20 ms–1? 75 kJ

Work required to increase KE• Ideally, the work done on an object is all

transferred to the increase in KE.– e.g. a 40 kg block of ice is pushed across a

smooth floor with a force of 100 N for 5 m. what is its final velocity?

Work done (F × d) = change in KE (½mv2) so v2 = 2Fd/m = 2×100×5 / 40 So v = 5 m/s

• Note that this assumes no energy is lost to friction or air resistance – in real life this is never the case and the final velocity will be smaller.

Potential Energy

• Potential energy (PE) is the energy stored in an object when you raise it up against the force of gravity.

• Energy change = work done

So PE gained = work done raising object

PE change = force × distance

=(mg) × h

• So change in PE = mgh

Potential energy

• A roofer carries 20 kg of tiles up a 10 m ladder to the roof. – What is the gain in PE of

the tiles?– How much work did the

roofer do lifting them?

2 kJ

2 kJ

– He accidentally drops one 500 g tile. How much PE does it lose as it falls to the ground? 50 J

Conservation of energy

• “Energy cannot be created or destroyed, it can only be transferred from one form to another”

• What energy transfers are taking place in the picture?

chemical → kinetic →potential

potential → kinetic

• When an object falls, PE is converted to KE. – e.g. a 20 kg cannon ball is dropped from the top

of the Eiffel tower, which is 320 m high. What is the maximum speed of the ball as it hits the ground?

loss of PE (mgh) = gain in KE (½mv2)

So v2 = 2mgh/m = 2×20×9.81×320 / 20

so v = 80 ms–1 (1sf)• Why will the actual final speed be less than this?

– Wind resistance will limit it

Objects falling due to gravity

PE and KE can interchange• E.g. on a rollercoaster this can happen

several times…

loss of PE (mgh) = gain in KE (½mv2)

v2 = 2mgh/m = 2gh = 2×9.81×50

so v = 31.6 m/s

In fact, work done to overcome friction and air resistance = mgh – ½mv2

Qs on p. 152

Power

• Power is the rate of transfer of energy– (or the rate of doing work, therefore)

units: watt (W)

1 W = 1 Js–1t

W

t

EP

Motive power

• When a powered vehicle moves at constant speed:

• Work done/s = force × distance/s = Fv• Constant speed so resistive forces = motive force

• When a powered vehicle gains speed• Motive power = gain in KE/s + energy lost to

resistance/s• Accelerating so motive force > resistive forces

Efficiency

• No machine which converts energy from one form to another is 100% efficient– Some energy is always “lost”– This is often due to friction of some kind– “wasted” energy (all energy?) tends to end up

as heat

powerinput poweroutput

machine tosuppliedenergy machineby done work

machine tosuppliedenergy totalmachineby nsferredenergy tra useful

Efficiency

• Now try questions on pp.160–1