Q.1 Fill in the blank and rewrite the completed statements:
1.1 kWh = J
Ans 1 kWh = 3.6 x 10^{6} J
2. Work is equal to the product of …………… .
Ans Work is equal to the product of force and displacement.
3. If the velocity of the body increased by 2 times, then kinetic energy will increase by …………… .
Ans If the velocity of the body increased by 2 times, then kinetic energy will increase by 4 times.
4. When a stone is thrown in the upward direction, the work done by the person throwing the stone is …………… .
Ans When a stone is thrown in the upward direction, the work done by the person throwing the stone is positive.
5. was invented by James Watt.
Ans Steam engine was invented by James Watt.
6.The kinetic energy of a body of mass 10 kg moving with a velocity of 2 m/s is …………… .
Ans The kinetic energy of a body of mass 10 kg moving with a velocity of 2 m/s is 20 J.
7 The velocity of a moving body is increased four times. Hence, its kinetic energy will be the initial
kinetic energy.
Ans The velocity of a moving body is increased four times. Hence, its kinetic energy will be 16 times the initial kinetic energy.
8.In MKS system the unit of work is …………… .
Ans In MKS system the unit of work is the joule.
9.Water stored in the water tank possesses ……………
Ans Water stored in the water tank possesses potential.
10.Units of work and are the same.
Ans Units of work and energy are the same.
11.When an object of mass 5 kg is lifted through 3 m by the application of the force, the work done by the applied force is …………… .
Ans When an object of mass 5 kg is lifted through 3 m by the application of the force, the work done by the applied force is 147J.
12. When an aircraft takes off, work is done by the gravitational force.
Ans When an aircraft takes off, negative work is done by the gravitational force.
13. Flowing water has energy.
Ans Flowing water has kinetic energy.
14. When 10 N force is applied on the body of mass 2 kg, it displaces through 3 m. In this case joule of
work will be done.
Ans When 10 N force is applied on the body of mass 2 kg, it displaces through 3 m. In this case 30 joule of work will be done.
15. 1 horse power is equivalent to Watt.
Ans 1 horse power is equivalent to 746 Watt.
16. “The revolution of the moon around the earth”. Here, the work done by the gravitational force exerted by the earth is zero because the displacement of the moon is at to the direction of the earth’s gravitational
force.
Ans “The revolution of the moon around the earth”. Here, the work done by the gravitational force exerted by the earth is zero because the displacement of the moon is at right angles to the direction of the earth’s gravitational force.
17. The unit of energy in the CGS system is …………… .
Ans The unit of energy in the CGS system is erg. 18 …………… erg = 1 joule.
Ans 10^{7} erg = 1 joule.
19 A freely falling body conserves its throughout the fall.
Ans A freely falling body conserves its total energy throughout the fall.
Q.2. Find the odd one out :
1. A raised hammer, water at a height, a moving ball, a stretched rubber band.
Ans A moving ball.
2. A moving ball possesses kinetic energy whereas the rest possess potential energy.
3.A moving ball, a stretched rubber band, a raised hammer, water at a height. Ans A moving ball.
3. A moving ball possesses kinetic energy whereas the rest possess potential energy.
dyne – cm, kilowatt, joule, erg.
Ans Kilowatt.
4.Kilowatt is a unit of power whereas the rest are units of energy.
Q. 3. Find corelated terms:
1. Work done is positive : θ = 0^{o} :: work done is negative : ……………
Ans Work done is positive : θ = 0^{o} :: work done is negative : θ = 180^{o}.
2. At θ = 0^{o} work done is positive. Similarly, work done will be negative at θ = 180^{o}.
3. Work : joule :: power : …………….
Ans Work : joule :: power : watt.
Joule is the unit of work. Similarly, the unit of power is watt.
4. 10^{7} erg : 1 joule :: 1 kWh : ……………
Ans 10^{7} erg : 1 joule :: 1 kWh : 3.6 × 10^{6} joule. Conversion of 10^{7} erg in joule is 1 joule. Similarly, 1 kWh is equal to 3.6 × 10^{6} joule.
Q .4. Match the pair :
1
Column “A” 
Column “B” 
i. Kinetic energy 
a. mgh 
ii. Work 
b. ½ mv2 
iii. Potential energy 
c. W / t 
iv. Power 
d. F. s 
e. mg 
Ans
i. Kinetic energy 
½ mv2 
ii. Work 
F. s 
iii. Potential energy 
mgh 
iv. Power 
W / t 
2
Column “A” 
Column “B” 
i. Muscular force 
a. Iron nails sticking to a magnet, toy magnet sticking to the door of a refrigerator 
ii. Mechanical force 
b. Lifting of book, An ox pulling a cart 
c. Vehicle stopping when brakes are applied, rubbing palms against each other 

d. Drill Machine, sewing machine 
Ans
i. Muscular force 
Lifting of book, An ox pulling a cart 
ii. Mechanical force 
Drill Machine, sewing machine 
3
Column “A” 
Column “B” 
a. Iron nails sticking to a magnet, toy magnet sticking to the door of a refrigerator 

i. Magnetic force 

ii. Gravitational force 

b. Hairs getting attracted to rubber balloon 

c. Small paper bits sticking to plastic comb 

d. A ball thrown up falls back to the ground, Glass when dropped falls to the floor 

Ans
Iron nails sticking to a magnet, toy magnet sticking to the door of a refrigerator 

i. Magnetic force 

ii. Gravitational force 
A ball thrown up falls back to the ground, Glass when dropped falls to the floor 
Q.5. State True or False :
1. As the object moves downwards, its kinetic energy goes on decreasing.
Ans False – As the object moves downwards, its kinetic energy goes on increasing.
2. Like humans, machines also require energy to do work.
Ans Like humans, machines also require energy to do work. – True
3.Energy is a cause of work.
Ans Energy is a cause of work. – True
4. Work done on an object depends on time.
Ans False – Work done on an object does not depend on time.
5.The swinging pendulum of a clock has kinetic as well as potential energy.
Ans The swinging pendulum of a clock has kinetic as well as potential energy. – True
6. A body at rest may possess energy.
Ans A body at rest may possess energy. – True
7.Potential energy of an object is independent of height of an object.
Ans False – Potential energy of an object depends upon the height of an object.
8.More energy is required for throwing a stone upwards than throwing it downwards.
Ans More energy is required for throwing a stone upwards than throwing it downwards. – True
9. W = F s Cosθ is the formula used to calculate the amount of work done when displacement is not in the direction of the applied force.
Ans W = F s Cosθ is the formula used to calculate the amount of 9. work done when displacement is not in the direction of the applied force. – True
10. The unit of energy in the MKS system is joule.
Ans The unit of energy in the MKS system is joule. – True
11. When force applied on a body moves it through some distance, work is said to be done.
Ans When force applied on a body moves it through some distance, work is said to be done. – True
12. Work is a scalar quantity.
Ans Work is a scalar quantity. – True
13. Energy is destroyed when it is transformed from one form to another.
Ans False – Energy is neither created nor destroyed when it is transformed from one form to another.
14. If a person is throwing some object, work done by him is negative.
Ans False – If a person is throwing some object, work done by him is positive.
15. As energy of the body increases, its power decreases.
Ans False – As energy of the body increases, its power increases.
Q.6.Name the following :
1. Force applied by an ox to pull a cart.
Ans Electrostatic force
2. The unit of work in the CGS system. Ans erg.
3. Force used when brakes are applied to stop vehicles.
Ans Frictional force
4. The energy in Flowing water.
Ans kinetic energy
Q.7. Multiple Choice Questions :
1. The work done on an object does not depend on …………… .
a. displacement
b. applied force
c. initial velocity of the object
d. the angle between force and displacement.
Ans Option c.
2. Three friends A, B and C start journey to reach same destination. A used his car, B uses his cycle and C goes walking who according to you is doing more work.
a. A and B b. B and C
c. only A d. All are doing same work
Ans Option d.
…………… are the forces involved in dragging a heavy object on a smooth, horizontal surface, having the same magnitude.
a. Horizontal applied force and gravitational force.
.b. Gravitational force and reaction force in the vertical direction
c. Horizontal applied force and reaction force in the vertical direction.
d .Gravitational force and force of friction.
Ans Option b.
3. Power is a measure of the ………….. .
a. The rapidity with which work is done
b.Amount of energy required to perform the work
c.The slowness with which work is performed
d.length of time
Ans Option a.
4. Loud speaker creates noise pollution. State the transformation of energy taking place in it – ……………
a. mechanical energy to sound energy b. sound energy to electrical energy
c. electrical energy to sound energy d. electrical energy to heat energy
Ans Option c.
5. Write an expression for work done by constant force acting on a body and displacement is in the direction different from the direction of force.
a. W = Fs b. W = F cosq
c. W = S cosq d. W = FS cosq
Ans Option d.
6. The potential energy of your body is least when you are …………….
a. sitting on a chair. b. sitting on the ground.
c. sleeping on the ground. d. standing on the ground.
Ans Option c.
7.If we increase the velocity of a car moving on a flat surface to four times its original speed, its potential energy …………….
a. will be twice its original energy. b. remains unchanged.
c. will be 4 times its original energy. d. will be 16 times its original energy.
Ans Option b.
8. Joule is the unit of …………….
a. force b. work c. power d. both (b) and (c)
Ans Option d.
9. The total energy of an object falling freely towards the ground …………… .
a. decreases b. remains unchanged.
c. increases. d. increases in the beginning and then decreases.
Ans Option b.
10. For work to be performed, energy must be ……………
a. Transferred from one place to another b. Concentrated
c. Transformed from one type to another d. Destroyed
Ans Option c.
11. While dragging or lifting an object, negative work is done by ……………
a. the applied force. b. frictional force.
c. gravitational force. d. both (b) and (c).
Ans Option d.
12. Coolie is standing with load on his head. Work done by him is zero because.
a. load is less
b.there is no displacement
c. load is on his head and not on back
d. All of the above
Ans Option b.
13. When the angle between force and displacement as 0° work done is – ……………
a. minimum b. maximum
c. negative d. All of the above
Ans Option b.
14. Give an example for transformation of electrical energy into chemical energy.
a. Secondary Cell b. Electric lamp
c. Primary Cell d. Engine
Ans Option a.
15. Which of the forces involved in dragging a heavy object on a smooth, horizontal surface, have the same magnitude?
a. The horizontal applied force b. Gravitational force
c. Reaction force in vertical direction d. Force of friction
Ans Option d.
Q.8. Solve Numerical problems :
1. Determine the amount of work done when an object is displaced at an angle of 30° with respect to the direction of the applied force.
Ans Given : Angle (θ) = 30°.
To find : Work done, W= ?
Solution :Work done on a body,
W= Force(F) x Displacement(s) x Cos θ. W = F s Cos θ
W = F s Cos 30° W = F s^{√3}/_{2}
W = ^{√3}/_{2} F s joule.
Therefore, the work done by the force is ^{√3}/_{2} F s joule.
Show that 1 kWh = 3.6 x 10^{6} J.
Ans Power is said to be of 1 kilowatt, when 1000 joules of work is done in 1 second.
1 kWh 
= 
1 kW x 1 hr 
= 
1000 W x 3600 s. 

1 kWh 
= = 
3600000 J. 3.6 x 10^{6} J. 
2. What is the kinetic energy of a body of mass 10gm moving with a velocity of 200m/s?
Ans Given : Mass, m = 10gm = 0.01kg Velocity, v = 200m/s
To find : Kinetic energy, K.E. = ?
Solution : Kinetic energy of a body is given by, K.E. = ½ x mass(m) x (velocity)²(v²)
K.E. = ½ x m x v²
K.E. = ½ x (0.01) x (200)²
K.E. = 200 J.
Therefore, the kinetic energy of the body is 200 J.
What will be the work done, if a body of mass ‘m’ is raised from ground to a height ‘h’?
Ans i. Weight of the body = mg
Gravitational force acting on the body, F = mg
3. The magnitude of the force required to lift the body is equal to its weight.
iv work done in raising the body against gravitational force
= force x displacement
= mg x h = mgh.
If an object has zero momentum, does it have kinetic energy? Explain your answer.
Ans i. Momentum(p) of an object is the product of its mass(m) and velocity(v) i.e. p = m x v
If momentum of an object is zero, then its velocity (v) is also equal to zero. i.e. v = 0
Kinetic energy of an object is given by, K.E. = ½ mv^{2} = ½ m (0)^{2} = 0
Therefore, if an object has zero momentum, it does not have any kinetic energy
If a 1200 W electric iron is used daily for 30 minutes, how much total electricity is consumed in the month of April?
Ans Given : Power (P) = 1200 W = 1200 x 10^{–3} kW. Time (t) = 30 min = 0.5 hr.
Number of days = 30 days(April).
To find : Electricity consumed(E) = ?
Solution : E = P x t x Number of days E = 1200 x 10^{–3} x 0.5 x 30
E = 18 unit.
Therefore, the total electricity consumed by electric iron is 18 unit.
4. Ravi applied a force of 10 N and moved a book 30 cm in the direction of the force. How much was the work done by Ravi?
Ans Given : Force, F = 10 N. Displacement, s = 30 cm = 30 x 10^{–2} cm. = 30 x 10^{–2} cm.
To find : Work done, W = ? Solution : Work done on a body, W = Force(F)xDisplacement(s) W = F x s
W = 10 x 30 x 10^{–2} m W = 3J Therefore, the work done by Ravi is 3 J.
4. If the energy of a ball falling from a height of 10 m is reduced by 40%, how high will it rebound?
Ans Given : Height (h_{1}) =10 m.
(P.E.)_{2 }=(P.E.)_{1} – 40% of (P.E.)_{1 ; }= 60% x (P.E.)_{1}
Therefore, the ball will rebound 6 m high.
6. Why does the K.E. of a body increases by nine times if the velocity of a body is increased by three times.
Ans i. The kinetic energy of a body of mass ‘m’ and velocity ‘v’ is given by, K.E. = ½ m v²
If the velocity of the body is now increased by three times, the new velocity will be ‘3v’.
Hence, the new K.E = ½ x m x (3v)² = ½ x m x 9v²
= 9 ( ½ mv² ).
Thus, Kinetic energy of a body increases by nine times if the velocity of a body is three times.
What is the potential energy of the body of 5kg, when it is lifted through 20m?
Ans Given : Mass, m = 5kg
Height, h = 20m
Acceleration due to gravity, g = 9.8m/s²
To find : Potential energy, P.E. = ?
Solution : Potential energy of a body is given by, P.E. = (m) x (g) x (h)
P.E. = m x g x h
P.E. = 5 x 9.8 x 20
P.E. = 980 J.
Therefore, the potential energy of the body is 980 J.
7. The velocity of a car increases from 54 km/h to 72 km/h. How much is the work done if the mass of the car is 1500 kg?
Ans Given : Initially velocity, u = 54 km/hr = 54 x^{ 5} = 15 m/s.
18
Final velocity, v = 72 km/hr = 72 x^{ 5} = 20 m/s.
18
Mass, m = 1500 kg.
To find : Work, W = ?
Solution :Work, W = Final K.E. – Initial K.E. W = ½ x m x v² – ½ x m x u² W = ½ x m ( v² – u²).
W = ½ x 1500 (400 – 225)
W = ½ x 1500 (175)
W = 131250.
Therefore, the amount of work done is 131250 J.
8. A force of 1N applied to the body displaces it through 1m in the direction of the force. Calculate the work done in ergs.
Ans Given : Force, F = 1N = 10^{5} dynes.
Displacement, s = 1m = 10² cm.
To find : Work done, W = ?
Solution : Work done on a body, W = Force(F) x Displacement(s)
W = F x s
W = 10^{5} dynes x 10² cm
W = 10^{7}
ergs.
Therefore, the work done on the body is 10^{7} ergs.
8. An electric pump has 2 kW power. How much water will the pump lift every minute to a height of 10 m?
Ans Given : Power (P) = 2 kW = 2 x 10^{3} W.
Time (t) = 1 min = 60 seconds.
Height(h) = 10 m.
To find : Mass (m) of water = ?
=
Solution : Power (P) =
Work (W) Time (t)
mgh t
2 x 103 = m×9.8×10
60
2×10^{3} ×60
m =
9.8×10
m = 1224.5 kg
Therefore, the amount of water lifted by the pump is 1224.5 kg.
Q.10. Attempt the following. 2
1 Define the term ‘Energy’ & state its units in MKS and CGS system.
Ans i. ‘Energy’ is defined as the capacity of a body to perform work.
Energy is a scalar quantity.
In the MKS system, the unit of energy is ‘joule’. In the CGS system, the unit of energy is ‘erg’.
Q.10. Give scientific reasons 10
1. A fast bowler takes a run–up before bowling a ball.
Ans i. When a fast bowler takes a run–up, he acquires greater kinetic energy.
This kinetic energy helps the bowler to deliver the ball at a greater speed.
Hence, a fast bowler takes a longer start while bowling.
2. While driving a nail into wood, the hammer is taken backward.
Ans i. When the hammer is taken back and then moved forward, it acquires greater velocity as compared with the velocity of just striking.
ii. As a result of the greater velocity, it has more kinetic energy and can drive the nail into the wood.
3. Force and displacement are vector quantities and work is a scalar quantity.
Ans i. Work is the product of magnitude of force and magnitude of displacement along the line of action of the force.
Thus, work possesses magnitude only and no direction.
Therefore, force and displacement both are vector quantities while work is a scalar quantity.
4. Wind rotates the blades of a windmill.
Ans i. Wind possesses kinetic energy.
ii. Hence, when it strikes the windmill, it uses this K.E. to rotate the blades of a windmill
5. Explosion of a bomb produces heat and light.
Ans i. The explosive mixture in a bomb possesses chemical energy.
ii. When the bomb explodes, the chemical energy inside the bomb is released in the form of heat, sound, light and kinetic energy.
Q.11. Give examples 2
1 Give four examples in which one type of energy is transformed into another type.
Ans Four examples in which one type of energy is transformed into another type are as follows :
In thermal power station heat energy is converted into electrical energy.
In case of green plants, during photosynthesis, solar energy is converted into chemical energy.
For the working of spaceships, solar energy is converted into electrical energy.
In Rocket engine or vehicle chemical energy is converted into kinetic energy.
Q. 12. Give explanation using the given statement: 9
1. Kinetic energy is never negative. Explain
Ans i. The kinetic energy of a body of mass ‘m’ and moving with a velocity ‘v’ is given by the formula : K.E. = ½ mv².
The square of a quantity is never negative, hence, v² is never negative.
The mass (m) of a body is also always positive.
Therefore, the kinetic energy, too, is never negative.
2. A moving car does not stop immediately when the brakes are applied, but it covers some distance before it stops. – Justify.
Ans i. When brakes are applied, a retarding force is exerted on the car in the opposite direction.
So, deceleration is produced.
A moving car possesses kinetic energy.
Some time is required for this kinetic energy to become zero due to conversion of this energy into heat energy and sound energy.
Therefore, a moving car does not stop immediately when the brakes are applied, but it covers some distance before it stops.
3. When the moon revolves around the earth, no work is done by the gravitational force exerted on the moon by the earth. Explain
Ans i. The moon revolves in a circular orbit round the earth. During the revolution of the moon around the earth, the gravitational force exerted by the earth on the moon acts along the radius of the orbit and the displacement of the moon is tangential to the circular orbit.
Therefore, the displacement of the moon is always perpendicular to the gravitational force exerted on it by the earth.
Hence, no work is done by the moon due to the gravitational force exerted by the earth on moon.
Q.13. Explain with the help of examples 15
1. Explain Positive work with 2 examples.
Ans Positive work : If the displacement of the body is in the direction of the force,work done by the force is said to be ‘positive’.
Examples :
When a body is traveling vertically downwards, it moves in the downward direction. In this case, the displacement of the body is in the direction of gravitational force exerted on it by the earth. Hence, work done by the body due to the gravitational force applied is positive.
When a batsmen hits a ball towards the boundary, the work done by the applied force is positive and the force and the displacement are in the same direction.
When a boy goes from first floor to third floor with his school bag, the work done by the boy due to the applied force is positive as the force & displacement are in the same direction.
2. Explain Negative work with 2 examples.
Ans Negative work : If the displacement of the body is opposite to the direction of the force, work done by the force is said to be ‘negative’.
Examples :
When a body is traveling vertically upwards, it moves in the upward direction. In this case, the displacement of the body is not in the direction of the gravitational force exerted on it by the earth. Hence, the work done by the body due to the gravitational force applied is negative.
When a ball is thrown upwards, then the displacement of the body is opposite to the force of gravity. Therefore, in this case, the work done by the force of gravity is negative.
When a cyclist applies brakes to his bicycle, yet the bicycle covers some distance before coming to a halt. In this case, the bicycle moves in the direction opposite to that of the force applied by the brakes. Hence, the work done due to the force applied by the brakes is negative.
3. Explain ‘Kinetic energy’ along with example.
Ans i. Energy possessed by a body by virtue of its motion is called its ‘Kinetic energy’.
Mathematically, K.E. = ½ mv² where K.E. = Kinetic energy, m = mass, v = velocity.
Kinetic energy is a scalar quantity.
In the MKS system, the unit of kinetic energy is joule while its CGS unit is erg. Examples :
A bullet moving with the high velocity possesses kinetic energy. Due to this energy, the bullet can penetrate into any object it strikes.
When a hammer hits the nail, its kinetic energy is transferred to the nail. Due to this energy, the nail penetrates into the wall.
4. Explain ‘Potential energy’ along with example.
Ans i. Energy possessed by a body due to its position or configuration is called its ‘Potential energy’.
ii. If ‘m’ is the mass of the body, the work done to raise the body to a height ‘h’ above the earth’s surface is equal to mgh.
.^{.}. P.E. = mgh. where P.E. = Potential energy, g = acceleration due iii.Potential energy is a scalar quantity.
In the MKS system, the unit of Potential energy is joule while its CGS unit is erg.
When the body is on the ground then h = 0.
Then potential energy of the substance on the earth is considered as zero.
Examples :
The work performed in stretching the spring is stored in the stretched spring and when we release the spring, the potential energy is used by the spring to regain the original shape.
When an arrow is pulled on a bow, the bow, the arrow, and the hands of the archer possess potential energy. So, the arrow moves with high velocity when it is released.
5. State the law of conservation of energy. Explain with examples.
Ans Law of conservation of energy states that :
Energy can neither be created nor destroyed.
It can be converted from one form to another.
The total amount of energy in the universe always remains constant.
Examples :
Water stored in the reservoir of a dam possesses potential energy. It is then allowed to flow under gravity, through large pipes. As water falls under gravity, its potential energy is converted to kinetic energy. When water falls on large turbines, this kinetic energy is used to rotate large turbines connected to electric generators. The generators start working and they generate electrical energy.
When we light a candle, chemical energy of wax gets converted into light and heat energy. Thus there is no production of new energy.
Q.14. Solve Numerical problems 15
1. Derive the formula for the kinetic energy of an object of mass ‘m’, moving with velocity ‘v’?
Ans i. Consider a stationary object (u = 0) of mass ‘m’ which is set into motion due to the applied force ‘F’.
Due to the applied force, the object undergoes an acceleration ‘a’ and after time ‘t’, the velocity of the object becomes ‘v’.
During this time‘t’, the object displaces through distance ‘s’.
vi. Hence, the work done on the object is given by W = F x s. (1)
According to Newton’s second law of motion, F = m x a. (2)
Similarly, according to Newton’s second equation of motion, s = ut + ½ a t^{2}.
s = 0 + ½ a t^{2}.
s = ½ a t^{2} (3)
Substituting values in eq. (2) and (3) in eq. (1), we get, W = ma x ½ a t^{2}
W = ½ m (a t)^{2} (4)
Using Newton’s first equation of motion, v = u + at. v = 0 + at = at.
Squaring on both sides, we get,
v^{2} = (at)^{2} (5)
From eq. (4) and (5), we get, W = ½ mv^{2}.
The kinetic energy gained by an object is equal to the amount of work done on the object.
i.e. K.E. = W
K.E. = ½ mv^{2}.
2. A stone of mass 0.5kg is released from the top of a tower of height 200m. Find the kinetic energy of a stone when it falls on the ground.
Ans Given : Mass, m = 0.5kg
Height, h = 200m
Acceleration due to gravity, g = 9.8m/s²
To find : Kinetic energy, K.E. = ?
Solution: For a freely falling body, Potential energy of the body at a certain height is equal to the Kinetic energy of the body on the ground.
Potential energy, P.E. = Kinetic energy, K.E. ──► (1)
Potential energy of a body at a certain height is given by, = mass(m) x acceleration due to gravity(a) x
P.E. height(h)
P.E. = m x g x h. ──► (2)
From equation (1) and (2), we have,
Potential energy, P.E. = Kinetic energy, K.E = m x g x h. Kinetic energy, K.E. = m x g x h
K.E. = 0.5 x 9.8 x 200
K.E. = 980 J
Therefore, the kinetic energy of the stone when it falls on the ground is 980 J.
3. Derive the formula for the work done when a body of mass ‘m’ is lifted through a height ‘h’? What is the potential energy at that position? When the body falls, what will be its kinetic energy before it touches the ground ?
Ans i. Let ‘m’ be the mass of the body.
Therefore the force required to lift it = mg.
Therefore the work done(W) in lifting the body to a height h = Force x displacement
W = mg x h = mgh.
This work done on the body is converted into potential energy.
Therefore the potential energy of the body when it is lifted, P.E. = mgh.
Again while falling down, the displacement(s) of the body in the downward direction is considered as negative.
Therefore s = – h.
Now, according to the kinematic equation of motion, v² = u² + 2as.
As the body is initially at rest, u = 0 m/s and as the acceleration due to gravity is directed downward, a = ─g.
v² = (0 m/s)² + 2(─g) (─h)
v² = 2 gh.
Therefore the K.E. of body, just before it touches the ground, K.E. = ½ mv²
K.E. = ½ m x 2gh = mgh.
4. Weight of an object A is 500N and that of B is 1000N. If the object A is lifted to a height of 10m, some work is done. Find the height through which the object B can be lifted, if the same amount of work is performed on it.
Ans Given : Weight of an object A, F_{A}
Weight of an object B, F_{B}
= 500N
= 1000N
Height upto which object A is lifted, s = 10m
Also, Work done on object A, W_{A} = Work done on object B, W_{B}
To find : Height upto which object B can be lifted = ?
Solution :When a body is lifted, force applied on the body is numerically equal to its weight.
In case of object A :
Work done on object A, W_{A}
= Force(F) x Displacement(s)
W_{A} = F_{A} x s
W_{A} = 500 x 10 W_{A} = 5000 J
In case of object B :
Also, Work done on object A, W_{A} = Work done on object B,W_{B}
W_{A} = W_{B}
W_{B} = 5000 J.
Work done on object B, W_{B} = Force(F) x Displacement(s)
W_{B} = F_{B} x s
5000 = 1000 x s
s = 5m
Therefore, the object B can be lifted to a height of 5m.
5. Prove that the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.
Ans i. Consider an object initially at point ‘A’, with potential energy = mgh.
When the object is displaced from ‘A’ to ‘C’, under the influence of gravity, it is said to be in free fall.
Let the velocity of the object at ground i.e., at point ‘C’ be v_{C}.
The initial velocity of the object is zero, distance covered by the object is the height of free fall (h) and acceleration of the object is due to gravity (g).
According to Newton’s third equation of motion, v^{2}
2
v
C
2
v
C
= u^{2} + 2 as.
= 0 + 2 gh.
= 2 gh. (1)
Kinetic energy of the object is given by, K.E. = ½ mv^{2}.
K.E. = ½ m v_{C}^{2}.
K.E. ½ m (2gh). [From (1)]
K.E. = mgh
Thus, the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.
Q.15. Complete the sentences in paragraph 3
1. Complete the paragraph:
(slower, work, displacement, force, potential energy, kinetic energy, energy, faster, power, upward)
A moving object can do work. An object moving does more work than an identical object moving
relatively …………… . Work can be defied as the product of force and …………… . Work is a form of …………… .
A speeding car possesses …………… while energy stored in winding the key of a toy car is …………… .
Ans A moving object can do work. An object moving faster does more work than an identical object moving relatively slower. Work can be defied as the product of force and displacement. Work is a form of energy. A speeding car possesses kinetic energy while energy stored in winding the key of a toy car is potential energy.
Q.16. Answer the following 18
1. Explain the correlation between work and energy.
Ans i. When an object is lifted, it is required to do work against the gravitational force exerted by the earth on the object.
Some amount of energy is lost while doing this work & object gains an equal amount of energy.
So, there is a transfer of energy during the work.
Examples :
When one kicks a football, he performs some work on the ball. In this process, the person loses a part of his energy and the ball gains an equal amount of energy.
When one lifts a stone, he has to do work against the earth’s gravitational force on the stone. In this process, person loses a part of his energy & the stone gains an equal amount of energy.
2. What is ‘Power’? State its unit.
Ans i. Power is defined as the rate at which work is performed.
Mathematically, Power( P ) = Work( W )
Time( t )
Power is a scalar quantity.
The units of Power in M.KS system: Joule /second( Watt ),
in C.G.S system: Erg /second,
Industrial unit: Horsepower. [1 horsepower = 746 watt]
3. When a bullet fired from the gun strikes the target, the target becomes hot.
Ans i. A bullet fired from a gun is moving with a high velocity.
Due to this energy it possesses high kinetic energy.
When it strikes the target, part of its kinetic energy is transformed into heat energy due to which the target becomes hot.
4. In order that a watch should work, it is necessary to wind up the spring from time to time.
Ans i. Potential energy is stored in the spring by winding it up.
The watch keeps on working due to the conversion of P.E. stored in the spring into K.E.
After some time potential energy becomes zero and the watch stops working.
Hence, it is necessary to wind up the spring of the watch from time to time.
5. Why does a driver increases the velocity of his car, when the car approaches an uphill road.
Ans i. When a car runs on a horizontal road, it has to do work only against the forces of friction and air resistance.
When the car moves on an uphill road, it has to do work against the gravitational force due to the earth in addition to the air resistance and friction.
Hence, in order to impart greater K.E. to his car, the driver increases the velocity of the car so that it can counter the combined effect of the forces of gravity, friction and air resistance.
6. Explain why does a blacksmith uses a hammer heavier than the one used by a goldsmith.
Ans i. A blacksmith has to beat plates from iron while a goldsmith prepares ornaments from soft metals like gold and silver.
A blacksmith is required to give more kinetic energy to his hammer than the goldsmith.
As the kinetic energy possessed by a body increases with the mass of the body, the blacksmith uses a hammer heavier than the one used by a goldsmith.
Q.17. Extra data (Not to be Use) 15
1. Give two examples in which one type of energy is transformed into another type.
Ans Two examples in which one type of energy is transformed into another type are as follows :
In thermal power station heat energy is converted into electrical energy.
In case of green plants, during photosynthesis, solar energy is converted into chemical energy.
2. The work done on an object moving with uniform circular motion is zero.
Ans 1. When an object is moving in uniform circular motion, the force acting on it is along the radius of the circle
and its displacement is along the tangent of the circle.
Thus they are perpendicular to each other.
So when the force and the displacement are perpendicular to each other, the work done by the force is zero.
Hence, The work done on an object moving with uniform circular motion is zero.
3. . Why does the work and energy have same units?
Ans i. Energy is the capacity of a body to perform work.
Therefore, the energy possessed by the body is measured in terms of the work done by it.
Hence, the work and energy have the same units.
Q.18. Answer the following in detail 25
1. What do you understand by unit work? Derive the units of work in the MKS (SI) system and CGS system and derive the relation between them.
Ans i] A unit work is said to be done when a unit force produces a unit displacement of a body along the line of action of the force.
ii] Mathematically, Unit work(W) = Unit force(F) x Unit displacement(s). i.e. if F = 1 unit and s = 1 unit, then W = 1 unit.
iii] It is a scalar quantity.
iv] Its MKS unit is joule while its CGS unit is erg.
Joule:
1] When a force of 1 newton displaces a body through 1 metre along the line of action of the force, the work done by the body is said to be 1 joule.
2] Joule is the unit of work in the MKS system.
3] Thus, 1 joule = 1 newton x 1 metre.
Erg :
1]When a force of 1 dyne displaces a body through 1 centimetre along the line of action of the force, the work done by the body is said to be 1 erg.
2.Erg is the unit of work in the CGS system.
3. Thus, 1 erg = 1 dyne x 1 centimetre.
Relation between Joule & erg : 1 Joule = 1 newton x 1 metre.
= 10^{5} dyne x 10^{2} cm
= 10^{7} dynecm
= 10^{7}
ergs.
2. Write the general expression for work in terms of force, displacement & angle between two. Hence, write the conditions under which work done is positive, zero, negative?
Ans
i]Let a constant force of magnitude ‘F’ acts on a block and displace it through a distance AB.
ii]Here, the displacement of the block is not in the direction of the force acting on the block.
iii]In this case, the angle between the direction of the force and that of the displacement is ‘θ’.
iv]For determining the displacement of the block in the direction of the force, from Y draw a perpendicular
BC along the line of action of the force.
v]Thus, AC will be the component of F in the direction of the displacement. Let AC = F_{1}.
vi]But, Cos θ =
AC = F1
AB F
Therefore, F_{1}
= F cos θ
vii]Thus, in this case, work done is given by,
W = → × =
\ viiiThe work done is positive for 0 ≤ θ < 90°.
ix]The work done is zero for θ = 90 X ]The work done is negative for 90° < θ ≤ 180°.
3. State the conditions under which no work is performed by a force applied on a body.
Ans The conditions under which no work is performed by a force applied on a body are as follows :
a. When the applied force produces no displacement : When the applied force does not displace the body, no work is performed by the body due to the applied force.
For example :
b. When the displacement of the body is perpendicular to the applied force : When a force is applied to a body and if the displacement of the body is perpendicular to the applied force, work done by the body due to the applied force is zero.
For example :
a.. Electrons are revolving around the nucleus. The displacement of the electron is perpendicular to the force applied by the nucleus on the electron.
1) The moon revolves in a circular orbit round the earth.
2) During the revolution of the moon around the earth, the gravitational force exerted by the earth on the moon acts along the radius of the orbit & the is tangential to the circular orbit.
3) Therefore, the displacement of the moon is always perpendicular to the gravitational force exerted on it by earth.
4) Hence, no work is done by the moon due to the gravitational force exerted by the earth on it
4. Define and explain the term ‘work’?
Ans A] Work is the product of the magnitude of force and magnitude of displacement along the line of action of the force.
B] Mathematically, Work(W) = Force(F) x displacement(s).
C] Work is a scalar quantity.
D] The unit of work in MKS system is joule.
E] The unit of work in CGS system is erg. ( Also, 1 joule = 10^{7}
Explanation :
ergs ).
A] When a force is applied to a body, the body moves in its own direction. Thus, in this case, work is said to be done by the body due to the applied force.
B] In other words, due to the application of force, if the displacement of the body takes place along the line of action of the force, work is said to be done by the body due to the applied force
Example :
I] Let a constant force of magnitude ‘F’ acts on a block & displace it through a distance ‘s’ in the direction of force, then the work done(W) is given by, W = F s.
II ] A labourer carries heavy loads from one place to another throughout the day. In this case, work is said to be done as some force results in displacement.
5. Study the following activity and answer the questions.
I] Take two alluminium channels of different lengths.
II] Place the lower ends of the channels on the floor and hold their upper ends at the same height.
III] Now take two balls of the same size and weight and release them from the top end of the channels. They will roll down and cover the same distance.
Questions
1] At the moment of releasing the balls, which energy do the balls have?
2] As the balls roll down which energy is converted into which other form of energy?
3] Why do the balls cover the same distance on rolling down?
4] What is the form of the eventual total energy of the balls?
5] Which law related to energy does the above activity demonstrate ? Explain.
Ans 1] At the moment of releasing the balls, which energy do the balls have?
At the moment of release, the balls will possess potential energy.
2] As the balls roll down which energy is converted into which other form of energy?
When balls roll down, potential energy gets converted into kinetic energy.
3] Why do the balls cover the same distance on rolling down?
Balls cover the same distance on rolling down because two balls are of same size and weight and are released from the same height. Therefore potential stored in them will be the same because potential energy depends only on the mass and height and is independent of the path followed.
4] What is the form of the eventual total energy of the balls?
The total energy fo the balls will be sum of their kinetic energy and potential energy. The total energy will always remain constant.
i.e., Total energy = Kinetic energy + Potential energy
5] Which law related to energy does the above activity demonstrate? Explain.
The above activity demonstrates law of conservation of energy correctly.
Initially when ball was at rest it had potential energy stored in ti. When the ball is released, its potential energy gets converted into kinetic energy and the net energy is conserved.