Prof. Muhammad Amin 1

Chapter-4 work and energy

Type equation here.

Q. 4.1 Describe the work done by a variable force.

Ans. Work Done by a Variable Force

We shall now consider the work done by a force which is not constant consider a

particle is moved from point ‘a’ to ‘b’ by the action of a variable force.

Let us divide the total displacement into n short intervals of

displacements 1 2Δd , Δd ,.....Δd ,n and 1 2 nF , F ,..........F ,are the

force acting during the intervals as shown in figure (a).

During each small interval, the force is supposed to be

constant.

Work done for the first interval is

11 1 1 1 1Δw =F Δd = F Cosθ Δd ____________(1)

Work done for the second interval is

22 2 2 2 2Δw =F Δd = F Cosθ Δd ____________(2)

Total 1 2 nW = Δw + Δw + ............+ Δw

1 1 1 2 2 2 n n n= F Cosθ Δd + F Cosθ Δd +..............+F Cosθ Δw

i=n

Total i i 1i=1

or W = FCosθ Δd _______________(2)

Work Done by Graphical Method

We can check the relation no. (2) By plotting the

graph between ‘F Cos’ and ‘d’ as shown in fig. (b)

The value of ‘F Cos’ at the beginning of each

interval is shown by the horizontal lines. In order to

have a better result, we can divide the total displacement for point ‘a’ to ‘b’ into very

large number of equal intervals, so that d tends to zero or n tends to infinity.

Thus equation no. (2) can be written as

i=n

Total i i iΔd 0

i=1

W = Lim FCosθ Δd

Q. 4.2 (a) Define gravitational field? Show that work done in a gravitational field is

independent of the path followed?

(b) Prove that the earth’s gravitational field is conservative field?

Ans. (a) Gravitational Field

The space around the earth in which its gravitational force acts on a body is called the

gravitational field.

Work done by the gravitational force

When an object is moved in the gravitational field the work is done by the gravitational

force.

Work and energy Chapter

4

Prof. Muhammad Amin 2

Chapter-4 work and energy

If displacement is in the direction of gravitational force, the work is positive. If

displacement is opposite to the gravitational force, the work is negative.

Consider an object of mass m is moved with constant velocity from point A to B along

the different paths in the presence of a gravitational force as shown in fig. (a).

In this case the gravitational force is equal to the weight W =m g of the object.

The work done by the gravitational force along the path ADB can be divide into two

parts.

1- The work done along AD is given by

ADW = w. h = -wh = -mgh

2- The work done along BD h is

W = w. h = -wh = -mghBD

owand hare in opposite direction, θ=180

ADB AD DBW = W W

ADBW = 0+ -mgh =-mgh____________(1)

Now consider the path ACB.

3- Work done along AC is.

W = -mghAC

wand h are in opposite directionAC

4- The work done along CB is.

W = 0CB

w is perpendicular to path CB

ACB AC CBW = W +W

= -mgh+0

= -mgh_____________(2)

Now consider the path 3, a curved path. Let it is divided into a large number of

horizontal and vertical steps as shown in fig (b).

Note There is no work done along the horizontal steps, because w = m g is perpendicular

to the displacement for these steps. Work is done by the force of gravity only along the

vertical displacements.

Prof. Muhammad Amin 3

Chapter-4 work and energy

ACB 1 2 nW = -mg Δy -mg Δy ...........-mgΔy

ACB 1 2 nW = -mg Δy + Δy +.........+Δy

1 2 nΔy + Δy +...........+Δy = h

ABW = -mg h ________________(3)

From equation 1, 2, 3, we can conclude that the net work done be the same.

(b) Conservative Field

A field in which the work done in moving a body along a closed path is zero is called

conservative field.

Consider an object of w = m g is moved along a closed path ACBA in the gravitational

field.The work done along AC is

WAC = – mgh

The work done along CB is

WCB = 0

As we know that work done along AB is

WAB = –mgh

WBA = + mgh

Thus WACBA = WAC + WCB + WBA

= – mgh + 0+ mgh

= 0

Q. 4.3 Define and derive a relation for absolute potential energy?

Ans. Absolute P.E

The absolute gravitational potential energy of an object at a certain position is defined

as the work done by the gravitational force in taking the object from that position to

infinity where the force of gravity becomes zero.

Prof. Muhammad Amin 4

Chapter-4 work and energy

Relation for Absolute Potential Energy

Consider a body is lifted from point I to a very large

distance at point N from the surface of earth in the

gravitational field. As gravitational force does not

remain constant between the points I and N. in order to

remove this difficulty, we divide the distance between the

points I and N in to large number of small steps, each of

length r. Since r is very small, the gravitational

force remains constant for each step.

Let us calculate the work done in moving a body from

point 1 to 2. If r1 and r2 are the distances of points 1 and 2 from the centre of the earth

as shown in fig

Then r2 – r1 = r _______________(1)

or r2 = r1 + r _______________(2)

If r is the distance between the centre of this part r and centre of the earth,

Then 1 2r +rr = ________________(3)

2

Putting the value of r2 from equation (2)

1 1 1r + r +Δr 2r +Δr

r = =2 2

12r Δror r = +

2 2

1

Δrr = r + ________________(4)

2

Taking square on both sides

2

21

Δrr = r +

2

22 2

1

Δrr = r + + 2

41

Δr×r ×

2

As 2Δr

4is very small quantity, so it is neglected.

2 21 1r =r +r Δr

Putting the value of r from equation (1)

2 21 1 2 1r = r + r r - r

2 21r = r 2

1 2 1+ r r - r

21 2r = r r ____________(5)

The magnitude of gravitational force between the body and the earth is given by

Prof. Muhammad Amin 5

Chapter-4 work and energy

2

mMF = G

r

Putting the value of r2 form equation (5)

1 2

mMF = G ___________(6)

r r

Where G = gravitational constant. M = mass of the earth

1 2W = F . Δr

. 180oF rCos

180oF rCos

= F Δr × -1 = - F Δr

Putting the values of F and r from equation (6) and (1)

1 2 2 1

1 2

W =GmM

r rr r

The negative sign shows that the work has to be done on the body from point 1 to 2

because displacement is opposite to gravitational force.

2 11 2

1 2

W =r r

GmMr r

=r

GmM 2

1r r

2

r 1

r

1 2r

1 2

1 2

1 1or W = GmM

r r

Similarly, the work done for other parts are

2 3

2 3

1 1W = GmM

r r

– – – – – – – – – – – – – – – – – – – – – – – –

– – – – – – – – – – – – – – – – – – – – – – – –

– – – – – – – – – – – – – – – – – – – – – – – –

1

2 1 3

1 1W =

N N

N N

GmMr r

Total work done is

1 2 2 3 1W = .............Total N NW W W

Putting their values and taking (– g m M) common.

Total

1 2

1 1W = - mM -

r rG

2

1+

r

3 N-1 N

1 1 1- +--+ -r r r

1 N

1 1= - mM - ___________(7)

r rG

Prof. Muhammad Amin 6

Chapter-4 work and energy

When the point N lies at an infinite distance from the earth’s surface, then

RN = (infinity) Putting in (7)

1

1 1- mM

rTotalW G

1

1 1= - mM - 0 0

rG

1

1- mM

rTotalW G

Therefore, the general relation for the gravitational potential energy of a body at a

distance r form the centre of earth is

mM

- _______________(8)G

Ur

Relation (8) is also known as the absolute value of gravitational potential energy of a

body at a distance r from the centre of the earth.

If the body is on the surface of earth, then

R = R (radius of the earth)

Putting in equation (8)

g

GmMU = -

R

According to definition, this is known as absolute potential energy on the surface of the

earth.

Q.4.4 Define escape velocity. Derive a relation for escape velocity and calculate its vale

on the earth’s surface?

Ans. Escape Velocity

The initial velocity of an object with which it goes out of the earth’s gravitational field,

is called escape velocity.

Relation for Escape Velocity

If a body of mass m is thrown upward from earth’s surface to a very large distance into

space such that it goes out of the earth’s gravitational field, then we must give it initial

K.E. equal to absolute P.E.

Initial K.E. = absolute P.E.

Prof. Muhammad Amin 7

Chapter-4 work and energy

1

2m 2

esc

M mV G

R

Where Vest. = escape velocity.

M = mass of the earth.

G = gravitational constant.

R = radius of the earth.

21

2esc

MV G

R

2 2esc

G Mor V

R

2 ________________(1)esc

GMV

R

2

MmAs F = G

R

Also F = w = mg

mM m

g=G2R

or gR2 = G m Putting in equation(1)

2

esc

2 gRV = = 2gR______________(2)

R

Value of Escape Velocity on the Earth Surface

6 -2R = 6.4 × 10 m, g = 9.8 ms

6escV = 2 9.8 6.4 10

3 1escV = 11 10 11ms or Km

Q. 4.5 (a) In the absence of air friction, show that energy of a falling body remains

constant?

(b) When air friction f is present?

Ans. Interco version of Potential Energy and Kinetic Energy

(a) Consider a body of mass m at rest, at height ‘h’ above the surface of earth.

Prof. Muhammad Amin 8

Chapter-4 work and energy

At the position A, the body has

P.E. = m g h _________(1)

K.E. = 0 _____________(2)

If the body is allowed to fall freely under the action of gravity, we calculate P.E. and K.E.

at the position B.

When the body is at B,

Then its height from the point C (just above the ground is (h – x).

P.E. = m g (h–x) _________(3)

and K.E. = 2B

1m v

2 _____________(3)

where VB = velocity at the position B.

i f BV= 0, V = V , S = x

Using the following relation

2 2BV = 0 +2gx = 2gx

Putting in equation (3)

1

K.E = m 2gx =mgx2

Total energy at B = P.E + K.E

= mg h-x +mgx

mgh m g x m gx

= mgh_____________(4)

At position c, just before the body strikes the earth, P.E. = 0

2C

1and K.E = mV

2

i f cV= 0, V = V , S = h

2cV =0+2gh = 2gh

1

K.E. = m 2gh2

K.E = mgh_______________(5)

Total energy at C= P.E. + K.E

= 0 + m g h

= m g h ______________________(6)

From equation (1), (4) and (6), we conclude that total energy remains constant at

positions A, B, and C when a body falls, its velocity increases. The increase in

velocity results in the increase in the increase in its K.E.

Prof. Muhammad Amin 9

Chapter-4 work and energy

If a body falls, its height decreases and hence, its P.E. also decreases.

From fig, we can write as loss in P.E = gain in L.E

2 21 2 2 1

1 1mgh -mgh = m V - m V

2 2

2 21 2 2 1

1mg h -h = m V -V

2

Where V1 and V2 are velocities of the body at the heights h1 and h2

(b) Let the air friction f is acting on the body during its fall. A part of P.E is used in dong

work against friction work done against frictional force is equal to f h.

The remaining P.E. = m g h – fh is converted into K.E.

21mgh -fh = m V

2

21mgh m V

2fh

Loss in P.E. = Gain in K.E> + work done against friction

Prof. Muhammad Amin 10

Chapter-4 work and energy

Q. 4.1 A person holds a bag of groceries while standing still, talking to a friend. A car is

stationery with its engine running. From the stand point of work, how are these

two situations similar?

Ans. in both cases work done is zero

Reason

The person is applying force against gravity in holding the bag. As displacement is

zero, so no work is done.

In case of stationary car also the work done is zero.

In both the cases, force is there but as displacement is zero

SO Work = F. d=0

Q. 4.2 Calculate the work done in kilo joules in lifting a mass of 10 kg (at a steady

velocity) through a vertical height of 10 m.

Ans. m = 10 kg

Height = d = h = 10 m

G = 9.8 m s–2

Work done = F d ( F = w = mg)

= m g h

= 10 9.8 10 = 980 J

Or work done = 980

1000KJ = 0.98 KJ

Q. 4.3 A force F acts through a distance L. the force is then increased to 3 F, and then

acts through a further distance of 2 L. Draw the work diagram to scale.

Ans. Take the displacement along x–axis and force along y–axis

As area under the force displacement graph gives the work done so

W = (F) (L) + (3F) (2L)

W= 7 F L

Important short questions

3L0 L 2L

F

2F

3F

L

F

x axis

y axis

Prof. Muhammad Amin 11

Chapter-4 work and energy

Q. 4.4 In which case is more work done? When a 50 kg bag of books is lifted through 50

cm, or when = 50 kg crate is pushed through 2m across the floor with a force of 50

N?

Ans. CASE-1 m = 50 Kg

h = 50 cm = 0.5 m

g = 9.8 m s–2 ( F = w = mg)

Work done = w = F h

w = m g h

w = 50 9.8 0.5 =245J

CASE-2 m = 50 Kg, F = 50 N d = 2m

Work done = w = Fd

w = 50 2 = 100 J

Result:- more work is done in case-1

Q. 4.5 An object has 1 J of potential energy. Explain what does it mean?

Ans. Since energy is defined as the ability to do work, so an object has 1 J of P.E means it

has ability to do work of 1 J

Example

Since P.E = (mg) (h)

⇒ 1 J = (1N ) (1 m)

Hence if an object of weight 1 N is raised to a height of 1 m, then object has energy of 1 J.

Q. 4.6 When a rocket re-enters the atmosphere, its nose cone becomes very hot. Where

does this heat energy come from?

Ans. Heat energy comes from the conversion of K.E into heat

Explanation:-

When a rocket re-enters the atmosphere, the air friction opposes its motion. Some of

the K.E. is used in doing work against air friction which is dissipated in the form of

heat there form, nose cone of rocket becomes hot.

Q. 4.7 What sort of energy is in the following

(a) Compressed spring

(b) Water in a high dam

(c) A moving car

Ans. (a) Elastic P.E. is stored in a compressed spring.

(b) The gravitational P.E is stored in a high dam.

(c) A moving car has kinetic energy.

Q. 4.8 What is meant by work done by a constant force?

Ans The work done on a body by constant force is defined as the product of the magnitude of

the displacement and the component of the force in the direction of displacement.

We define work done w by the force F as the scalar product of F and d

. cos cosW F d Fd F d

The quantity cosF is the component of the force in the direction of the displacement

d .

Prof. Muhammad Amin 12

Chapter-4 work and energy

Q. 4.9 A girl drops a cup form a certain height, which breaks into pieces. What energy

changes are involved?

Ans. At a certain height, the cup has potential energy. When it is falling, P.E is converted

into kinetic energy. When the cup breaks in to pieces after hitting the ground, the K.E.

is converted into heat and sound.

Hence list of possible energy changes are:-

Potential energy

Kinetic energy

Work done in breaking the cup

Sound energy

Kinetic energy of pieces of cup

Heat energy

Q. 4.10 A boy uses a catapult to throw a stone which accidentally smashes a greenho use

window. List the possible energy changes.

Ans. A stone has elastic P.E. before throwing.

The following are the energy changes.

When a boy throws a stone by a catapult, the elastic P.E is converted into kinetic

energy.

When the stone strikes the window, its K.E. is converted into work done in

smashing the window.

Some of the K.E. of the stone is converted in to sound and heat.

A part of K.E. of the stone is lost against friction of air.

The remaining energy of stone is taken by broken pieces of window as a form of

K.E.

Q. 4.11 Under what condition that work done on a body has its positive value and negative

value?

Ans Work done is given by . cosW F d Fd

If the angle between force F and displacement d is less than 90o then in this case

cos will have positive value.

If the angle between force F and displacement d is greater than 90o then in this case

cos will have negative value.

Prof. Muhammad Amin 13

Chapter-4 work and energy

Q. 4.12 What does the area under the force - displacement graph represent?

Ans The area under the force – displacement graph represents

the work done on the body. If a body covers some

displacement d on the application of force F such that

are oriented at an angle . Then graph is drawn between d

along x – axis and along y – axis.

Area under graph = cosF (d)

= Work done

Q. 4.13 How can we calculate the work done in case of variable force?

Ans Work done by a variable force is computed by dividing the path

into very small displacement intervals such that during each

displacement, force remained nearly constant. Then calculate

the work done during each interval and take the sum of work

done for all displacement intervals which gives.

1

cosn

i i i

i

W F d

Q. 4.14 Differentiate between conservative and non-conservative forces, also give examples

of each.

Ans

Conservative Force Non Conservative Force

Work done by conservative force

is independent of path followed

Work done by conservative force

along a closed path is zero.

Examples of conservative forces

are gravitational force, elastic

spring force and electric force etc.

Work done by non conservative

force depends upon the path

followed.

Work done by non conservative

force along a closed path is not

zero.

Example of non conservative

forces are frictional force, air

resistance, tension in string,

normal force, propulsion force of

rocket or motor.

Q. 4.15 What is essential condition for conservative field?

Ans Conditions for conservative field

(i) In conservative field the work done is independent of the path followed.

(ii) Work done in a closed path must be zero in conservative field.

Q. 4.16 Define average power and instantaneous power and its unit.

Ans Average Power

It is the ratio of total work done to the total time taken by a body. If work w is done in

a time interval t then

Prof. Muhammad Amin 14

Chapter-4 work and energy

av

wP

t

Instantaneous Power

It is defined as the work done in an extremely small interval of time (approaching to

zero). 0

liminst

wP

t

S.I unit of power is watt.

Watt “It is defined as one Joule of work done in one second”

1

11

JW

S

Q. 4.17 A 70 kg man runs up a long flight of stairs in 4s. The vertical height of the stairs is

4.5m. Calculate power output in watts.

Ans m = 70kg

h = 4.5 m

t = 4sec

P =?

mgh

Pt

2

70 9.8 4.5

4

7.7 10P W

Q. 4.18 Write different methods to convert the solar energy into electrical energy?

Ans There are two methods.

Direct conversion

Photovoltaic cells convert solar energy to electrical energy.

Indirect Conversion

Solar energy heats up a tank of water and produces steam. Which is used to run steam

turbines and hence electricity is produced.

Q. 4.19 Describe some uses of solar cells?

Ans Solar cells are used

To power satellite, having large solar panels which are kept facing the sun.

In remote ground based weather stations.

In rain fore cast communication system.

In solar calculator

Q. 4.20 What are two common methods to convert biomass into energy?

Ans There are many methods used for the conversion of biomass into fuels. But the most

common are

(i) Direct combustion

Direct combustion of waste product like wood waste, crop residue and par ticularly

municipal solid waste.

Prof. Muhammad Amin 15

Chapter-4 work and energy

(ii) Fermentation

Fermentation of biomass using enzymes and by decomposition through bacterial action in

the absence of air (oxygen), by this process, we obtain Biofuel (ethanol) and biogas (a

mixture of methane and CO2)

Q. 4.21 Prove the relation .P FV

Ans Consider a constant force F acts on a body and it moves through a displacement d in

time t , then instantaneous power is given by

0

liminst

WP

t

We know that .W F d

0

0

0

.lim

= . lim

lim

t

t

t

F dP

t

dF

t

dSince V

t

Hence .P FV

Q. 4.22 What is meant by Non-conventional energy sources? Mention four of these energy

sources?

Ans Non-conventional energy sources

These energy sources are not very common these days. However, in future, these sources

will be used to meet the demands of energy.

(i) Solar (ii) Bio mass (iii) Fossil fuels (iv) Wind

Q. 4.23 Define work. Explain when it is maximum and when it is zero .

Ans Work The dot product of force and displacement is called work.

W = F.d

Maximum work

The work done has its maximum value, when force F and displacement d are parallel to

each other.

.

cos 0

cos 0

o

W F d

Fd

Fd

W Fd

Zero work

The work done has its minimum value, when force F and displacement d are

perpendicular to each other.

.

cos 90

cos 90

0

o

W F d

Fd

Fd

W

Prof. Muhammad Amin 16

Chapter-4 work and energy

Q. 4.24 What is kilo watt hour. Prove that 1kWh = 3.6 MJ.

Ans

Kilo watt hour is the commercial unit of electrical energy, it is defined as

“The amount of work done when a power of one kilowatt is maintained for one hour”

Proof:-

1kWh = 1000 W 3600 s.

1kWh = 3.6 106 watt sec

= 3.6 MJ ∵ 𝟏 𝐰𝐚𝐭𝐭 𝐬𝐞𝐜 = 𝟏 𝐉 𝐚𝐧𝐝 𝟏 𝐌 = 𝟏𝟎𝟔

Hence 1kWh= 3.6 MJ

NUMERICAL PROBLEMS

4.1 A man pushes a lawn mower with a 40 N force directed at an angle of 20o downward from the

horizontal. Find the work done by the man as he cuts a strip of grass 20 m long.

Solution:

Data:

Force exerted = F = 40N,

Length of strip of grass = d = 20m

Angle = 0θ = 20

To Find:

Work = W = ?

Calculation:

We know that

W = F.d

W = F d cos

= 40 (20) cos 200

= 800 x 0.93

W = 7.51 x 102J Ans.

4.2 A rain drop (m = 3.35 x 10-5 kg) falls vertically at a constant speed under the influence of the

forces of gravity and friction. In falling through 100 m, how much work is done by

(a) gravity and (b) friction.

Solution:

Data:

Mass of rain drop = m = 3.35 x 10-5 kg

Height = h = 100 m

To Find:

Work done by gravity = ?

Work done by friction = ?

Numerical Problems

Prof. Muhammad Amin 17

Chapter-4 work and energy

Calculations:

Work done by gravity = weight . height

= w.h = wh cos

Since weight and displacement are in the same direction, therefore, angle between them is 0o)

So,

W = Wh (1) cos0 1o

= Wh

W = mgh W mg

Substituting the given values, we have

W = 3.35 x 10-5 x 9.8 x 100

W = 0.0328

W = 0.0328 J Ans.

Work done by friction = f .h = fh cos180o = -fh cos180 1o

(Since friction is opposite to displacement, therefore, angle between them is 180o

As the drop is falling at a constant speed, the magnitude of force is equal to the weight of the drop hence

W = -f h = - mgh

= - 3.35 x 10-5 x 9.8 x 100

W = - 0.0328 J Ans.

4.3 Ten bricks, each 6.0 cm thick and mass 1.5 kg, lie flat on a table. How much work is required to stack them one on the top of another? Solution:

Data:

Height of each brick = h = 6cm = 0.06m

Mass of each brick = m = 1.5kg

Number of bricks = 10

To Find:

Work done by placing them one on the top of another = W = ?

Calculation:

No work is done for 1st brick.

Work done in placing other 9 bricks is stored in the form of P.E total work done to place one brick on the other = mg [ h + 2h + 3h + 4h + 5h + 6h + 7h + 8h + 9h]

= mgh [1+2+3+4+5+6+7+8+9]

= 45mgh

Prof. Muhammad Amin 18

Chapter-4 work and energy

= 45 x 1.5 x 9.8 x 0.06

= 39.69J

W = 40J App

4.4 A car of mass 800 kg traveling at 54kmh-1 is brought to rest in 60 meters. Find the average retarding force on the car. What has happened to original kinetic energy? Solution: Data:

Mass of the car = m = 800kg

Initial velocity = vi = 54 km h-1

1

54 1000

60 60

15

ms

Final velocity = vf = 0

Distance covered by car = d = 60

To Find: (i) Average retarding force = F = ?

(ii) What happened to original kinetic energy?

Calculation:

According to work energy principle

2 2

f i

1 1Fd mv mv

2 2

2 21 1

F 60 800 0 800 152 2

0 400 225F

60

F = -1500 N Ans.

The negative sign indicates that the force is retarding one whose magnitude is 1500 N.

4.5 A 1000 kg automobile at the top of an incline 10 metre high and 100 m long is released and

rolls down the hill. What is its speed at the bottom of the incline if the average retarding

force due to friction is 480 N?

Solution:

Data:

Mass of automobile = m = 1000 kg

f = 480 N

Length = = 100 m

To Find:

Final velocity = v = ?

Calculation:

We know that

21K.E. mv

2

Prof. Muhammad Amin 19

Chapter-4 work and energy

P.E. = mgh

According to law of conversion of energy

Loss of P.E. – work done against friction = Gain in K.E

So we can write

21mg h f mv

2

Substituting the value, we have;

1000 x 9.8 x 10 – 480 x 100 = 1

2 x 1000 x v2

98000 – 48000 = 500v2

50000 = 500 v2

100 = v2

v = 10 m s-1 Ans.

4.6` force (thrust) of 400 N is required to overcome road friction and air resistance in propelling

an automobile at 80kmh-1. What power (kW) must the engine develop?

Solution

Data:

Force required = F = 400 N

Velocity of automobile = v = 80 km/h

= 80×1000

3600

= 22.22 m/s

To Find:

Power = P = ?

Calculation:

We know that

P = F.v

P = F v cos

As force and velocity are in same direction so angle between them is 0o

oP Fvcos0

P = 400 x 22.2

= 8.9 x 103 W

P = 8.9 KW Ans.

4.7 How large a force is required to accelerate an electron (m = 9.1 x 10-31kg) from rest to a

speed of 2.0 x 107ms-1 through a distance of 5.0 cm?

Solution:

Prof. Muhammad Amin 20

Chapter-4 work and energy

Data:

Mass of electron m = 9.1 x 10-31 kg

Initial velocity = vi = 0

Final velocity = vf = 2 x 107 m/s

Distance = d = 5cm = 0.05m

To Find:

Force required = F = ?

Calculation:

Using work energy principle,

Work done = change in K.E

F d = 1

2 mvf

2 –1

2mvi

2

F d = 1

2 m (vf

2 – vi2)

2 231 71

F 0.05 9.1 10 2 10 0 )2

F(0.05) = 1

2 x 9.1 x 10-31 x 4 x 1014

17F 0.05 18.2 10

1718.2 10

F0.05

F = 3.6 x 10-15 N Ans.

4.8 A diver weighing 750 N dives from a board 10 m above the surface of a pool of water. Use

the conservation of mechanical energy to find his speed at a point 5.0 m above the water

surface, neglecting air friction.

Solution:

Data:

Weight of diver = W = 750N

h1 = 10m

h2 = 5m

h = h1 – h2 = 5m

2

2

v 2 9.8 10 5

v 2 9.8 5

v2 = 98

v 98

v = 9.9ms-1 Ans

To Find:

Speed of diver = v = ?

Calculation:

We know that

Loss in P.E = gain in K. E

2

1 2

2

1 2

1mgh mgh mv

2

1mg h h mv

2

2

1 2v 2g h h