The document Solution of Work and Energy (Page No - 145) - Physics by Lakhmir Singh, Class 9 Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Physics Solutions By Lakhmir Singh & Manjit Kaur.

All you need of Class 9 at this link: Class 9

__Page 145__**Solution 1**

Mass = m

Height above the ground = h

Work done = Potential energy acquired by the body = m x g x h

where g is acceleration due to gravity**Solution 2**

SI unit of work is Joule (J).**Solution 3**

Work is a scalar quantity.

Work, as a physical quantity, requires only magnitude to be represented. Hence, it is scalar quantity.**Solution 4**

When a force of 1 newton moves a body through a distance of 1 metre in its own direction, then the work done is known as 1 joule.**Solution 5**

The condition for a force to do work on a body is that it should produce motion in the body.**Solution 6**

Energy is a scalar quantity. It has only magnitude but no direction.**Solution 7**

a)Unit of work is joule.

b) Unit of energy is joule.**Solution 8**

The work done against gravity is zero when a body is moved horizontally along a frictionless surface because force of gravity acts perpendicular to the direction of motion.**Solution 9**

Kinetic energy will become four times when the speed is doubled because kinetic

energy is directly proportional to square of speed of the body

K.E ∝ V^{2}

**Solution 10**

Mass = m

Velocity = v

**Solution 11**

Kinetic

energy will become one-fourth when the speed is halved because kinetic energy

is directly proportional to square of speed of the body

K.E ∝ V^{2}

**Solution 12**

The kinetic energy of a body depends on

a) Mass of the body, m

b) Square of the velocity of the body, v^{2}**Solution 13**

Doubling the velocity would have a greater effect on kinetic energy.**Solution 14**

Mass = 50 kg

Kinetic energy = 625 J

**Solution 15**

a) Both kinetic and potential energy

b) Both kinetic and potential energy

c) Only kinetic energy

d) Only potential energy

e) Only potential energy

Note: In all the above cases we take ground as reference level where potential energy is zero.**Solution 16**

Let masses of body A and B be m

Height of body A = h

Height of body B = 2h

Potential energy for body A, PE_{A} = m x g x h

Potential energy for body B, PE_{B} = m x g x 2h

**Solution 17**

Mass = 1 kg

Velocity = 2 m/s

K.E.

**Solution 18**

Potential energy is a scalar quantity as it has magnitude only and it does not require any specification of direction.**Solution 19**

Mass = 100 kg

Height = 5 m

g = 9.8 m/s^{2}

P.E. = m x g x h = 100 x 5 x 9.8 = 4900 J

Work done is equal to PE acquired by the body.**Solution 20**

False.

PE = m x g x h

= 1 x 9.8 x 1 = 9.8 J**Solution 21**

The potential energy is doubled when the height is doubled since potential energy is directly proportional to height, h to which body is raised.

PE ∝ h

**Solution 22**

a) Potential energy

b) Kinetic energy

c) Potential energy

d) Potential energy

e) Potential energy**Solution 23**

a) Force ; distance

b) Zero

c) newton; metre; force

d) Energy; kinetic energy

e) mechanical**Solution 24**

The work done by a force on a body depends on two factors

a) magnitude of force

b) distance through which the body moves

Work done is directly proportional to the force applied and the distance through which the body moves.

W = F x s

where W is work done, F is force applied and s is distance through which the body moves.**Solution 25**

Yes, it is possible that a force is acting on a body but still work done is zero. For example, in the case of a man pushing a wall, the work done is zero despite of non-zero force, sin ce there is no displacement of the wall.

**Page No - 146****Solution 26**

a) Work done by force applied by the boy is positive because this force is in the direction of motion of the body.

b) Work done by the gravitational force is negative because this force is against the direction of motion of the body.**Solution 27**

W = F cos θ x s

where w is work done

F is force applied

θ is angle between the direction of force and the direction of motion of the body

s is the distance moved by the body

**Solution 28**

Kinetic energy is directly proportional to the mass of the body, m

KE ∝ m

Kinetic energy is directly proportional to the square of speed of the body, v

KE ∝ v^{2}

**Solution 29**

(a) Positive work: Work done by the force applied by a person on a ball that is thrown upwards.

(b) Negative work: Work done by gravitational force of earth on a ball thrown upwards.

(c) Zero work: Work done by gravitational force of earth on a box that is sliding horizontally on the ground.**Solution 30**

Mass 200 gm = 0.2 kg

Height = 5 m

intial velocity u = 0

Acceleration due to gravity = g = 9.8 m/s^{2}

final velocity , v

using third equation of motion

Kinetic energy =

Put the value of v^{2 }from eqn. 1

Kinetic energy =

**Solution 31**

**Solution 32**

Let mass of two object be m

v_{1} = 2m/s

v_{2}= 6 m/s

**Solution 33**

Mass of body = 2 kg

Initial velocity u = 0

Time taken = 2 s

Acceleration due to gravity, g = 10 m/s^{2}

Final velocity v

Using first equation of motion

v = u + gt = 0 + 10 x 2 = 20 m/s

**Solution 34**

Mass of scooter + scooterist = 150 kg

Initial velocity u = 10 m/s

Final velocity v = 5 m/s

Retardation = a

Distance covered= s

Using third equation of motion

v^{2} – u^{2} = 2as

5^{2} – 10^{2} = 2as

Work done W = F x s

But F = m x a

So, W= m x a x s

Put the value of ‘as’ from eq(i)

W = 150 x (-75/2) = -5625 J

Neagtive sign implies that force of brakes acts opposite to the direction of motion.**Solution 35**

Mass of rock = 10 kg

Height of ladder, h = 5 m

Initial velocity of rock, u = 0

Final velocity v

g = 10 m/s^{2}

using third equation of motion

v^{2} – u^{2} = 2gh

v^{2} – 0^{2} = 2 x 10 x 5

v = 10 m/s

**Solution 36**

Mass of car = 1000 kg

Initial velocity u = 20 m/s

Final velocity v = 10 m/s

Retardation = a

Distance covered = s

Using third equation of motion

v^{2} -u^{2} = 2as

10^{2}– 20^{2} = 2as

as = -150 ——(i)

Work done W = F x s

But F = m x a

So, W= m x a x s

Put the value of ‘as’ from equation (i)

W = 1000 x -150 = -150000 = -150 kJ Neagtive sign implies that force of brakes acts opposite to the direction of motion.**Solution 37**

Height , h= 10 m

Acceleration due to gravity, g = 10 m/s^{2}

i) Work done, W = m x g x h = 100 x 10 x 10 = 10000 = 10 kJ

ii) Potential energy of the body = work done = 10 kJ**Solution 38**

Height, h = 100 m

Acceleration due to gravity, g = 9.8m/s^{2}

Work done by the boy, W = m x g x h = 50 x 9.8 x 100 = 49000 J = 49 kJ

Potential energy gained by the boy = work done by the boy = 49 kJ**Solution 39**

Work done by a force applied on a body is

i) positive when the force acts in the direction of motion of the body.

ii) negative when the force acts in the direction opposite to the direction of motion of the body.

iii) zero when the force acts at right angle to the direction of motion of the body.**Solution 40**

Mass of the box, m = 150 kg

PE = 7350 J

Acceleration due to gravity, g = 9.8m/s^{2}

PE = m x g x h

7350 = 150 x 9.8 x h

**Solution 41**

Mass of the body, m = 2 kg

Initial velocity, u = 20 m/s

Acceleration due to gravity, g = 10 m/s^{2}

Height reached = h

Time, t = 2 s

Using second equation of motion

PE after 2 s = m x g x h = 2 x 10 x 20 = 400 J

**Solution 42**

Force, F = 1 N

Distance, s = 1 m

Work done W = F x s = 1 x 1 = 1 J**Solution 43**

Force, F = 2.5 x 10^{10} N

Velocity, v = 5 m/s

Time, t = 2 minutes = 120 s

Distance, s = v x t = 5 x 120 = 600 m

Work done, W = F x s = 2.5 x 10^{10} x 600 = 15 x 10^{12} J**Solution 44**

A stretched rubber band is an example of a body possessing energy while it is not in motion. The rubber band contains potential energy due to the change in its shape or configuration.**Solution 45**

a)Gravitational potential energy of a body depends on:

i)mass of the body, m

ii)height to which the body is lifted, h

iii)acceleration due to gravity, g

b)

i.A moving cricket ball has kinetic energy

ii.A stretched rubber band has potential energy**Solution 46**

Two examples where a body possesses both kinetic energy as well as potential energy are

i)a man climbing up a hill

ii)a flying aeroplane**Solution 47**

Mass of man, m

Height of tree, h = 5 m

Work done, W = 2500 J

Acceleration due to gravity, g = 10 m/s^{2}

W = m x g x h

2500 = m x 10 x 5

**Solution 48**

Work done, W =24.2 J

Distance, s = 20 cm = 0.2 m

Force, F

W = F x s

24.2 = F x 0.2

F = 24.2/0.2 = 121 N**Solution 49**

Mass of boy, m = 40 kg

Height, h = 1.5 m

Acceleration due to gravity, g = 10 m/s^{2}

i)At highest point, velocity, v = 0

Therefore KE = 0

ii)PE = m x g x h = 40 x 10 x 1.5 = 600 J**Solution 50**

a) Potential energy

b) Both potential and kinetic energy

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!