Chapter 10: Work and Energy Exercise & In-Text Questions & Answers
1A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force (Fig. 10.3). Let us take it that the force acts on the object through the displacement. What is the work done in this case?
Answer & Reference:
Given: Force (F) = 7 N, Displacement (s) = 8 m
Work done (W) = F × s = 7 N × 8 m = 56 J
(Page 115, In-text Question)
2When do we say that work is done?
Answer & Reference:
Work is done when two conditions are satisfied: (i) a force acts on an object, and (ii) the object is displaced in the direction of the force. (Page 114, Section 10.1.2)
3Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Answer & Reference:
Work done (W) = Force (F) × Displacement (s)
W = F × s (Page 115, Equation 10.1)
4Define 1 J of work.
Answer & Reference:
1 joule (J) is the amount of work done on an object when a force of 1 N displaces it by 1 m along the line of action of the force. (Page 115)
5A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?
Answer & Reference:
Given: Force (F) = 140 N, Displacement (s) = 15 m
Work done (W) = F × s = 140 N × 15 m = 2100 J
(Page 116, In-text Question)
6What is the kinetic energy of an object?
Answer & Reference:
Kinetic energy is the energy possessed by an object due to its motion. (Page 117, Section 10.2.2)
7Write an expression for the kinetic energy of an object.
Answer & Reference:
Kinetic energy (Ek) = ½mv²
where m is the mass of the object and v is its velocity. (Page 118, Equation 10.5)
8The kinetic energy of an object of mass, m moving with a velocity of 5 m s⁻¹ is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?
Answer & Reference:
Power is defined as the rate of doing work or the rate of transfer of energy. (Page 122, Section 10.3)
10Define 1 watt of power.
Answer & Reference:
1 watt is the power of an agent which does work at the rate of 1 joule per second. (Page 122)
11A lamp consumes 1000 J of electrical energy in 10 s. What is its power?
Answer & Reference:
Given: Energy consumed (W) = 1000 J, Time (t) = 10 s
Power (P) = W/t = 1000 J / 10 s = 100 W
(Page 123, In-text Question)
12Define average power.
Answer & Reference:
Average power is obtained by dividing the total energy consumed by the total time taken. (Page 122)
13An object of mass 20 kg is dropped from a height of 4 m. Fill in the blanks in the following table by computing the potential energy and kinetic energy in each case. (Take g = 10 m s⁻²)
Answer & Reference:
Total energy at height 4 m = mgh = 20 × 10 × 4 = 800 J
Using conservation of energy: PE + KE = 800 J at all points
(Page 122, Activity 10.15)
Height (m)
Potential Energy (J)
Kinetic Energy (J)
Ep + Ek (J)
4
800
0
800
3
600
200
800
2
400
400
800
1
200
600
800
Just above ground
0
800
800
14Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term 'work'.
• Suma is swimming in a pond.
• A donkey is carrying a load on its back.
• A wind-mill is lifting water from a well.
• A green plant is carrying out photosynthesis.
• An engine is pulling a train.
• Food grains are getting dried in the sun.
• A sailboat is moving due to wind energy.
Answer & Reference:
• Suma swimming: Work is done as she applies force to push water backward and moves forward.
• Donkey carrying load: No work is done in horizontal direction as force (upward) and displacement (horizontal) are perpendicular.
• Wind-mill lifting water: Work is done as force is applied against gravity to lift water.
• Photosynthesis: No work is done as there is no displacement of object by force.
• Engine pulling train: Work is done as force is applied in direction of displacement.
• Food grains drying: No work is done as there is no force causing displacement.
• Sailboat moving: Work is done as wind applies force in direction of boat's motion.
(Page 124-125, Exercise Q1)
15An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Answer & Reference:
Work done by gravity = mgh, where h is vertical displacement.
Since initial and final points are at same height, h = 0.
Therefore, work done by gravity = mg × 0 = 0 J.
(Page 125, Exercise Q2)
16A battery lights a bulb. Describe the energy changes involved in the process.
Answer & Reference:
Chemical energy in the battery → Electrical energy → Light and heat energy in the bulb.
(Page 125, Exercise Q3)
17Certain force acting on a 20 kg mass changes its velocity from 5 m s⁻¹ to 2 m s⁻¹. Calculate the work done by the force.
Answer & Reference:
Given: m = 20 kg, u = 5 m/s, v = 2 m/s
Work done = Change in kinetic energy = ½m(v² - u²)
= ½ × 20 × (2² - 5²) = 10 × (4 - 25) = 10 × (-21) = -210 J
(Page 125, Exercise Q4)
18A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.
Answer & Reference:
Work done by gravity = mgh, where h is vertical displacement.
Since movement is horizontal, h = 0.
Therefore, work done by gravity = 10 × g × 0 = 0 J.
(Page 125, Exercise Q5)
19The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Answer & Reference:
No, it does not violate the law of conservation of energy. As the object falls, its potential energy decreases but kinetic energy increases by the same amount. The total mechanical energy (PE + KE) remains constant.
(Page 125, Exercise Q6)
20What are the various energy transformations that occur when you are riding a bicycle?
Answer & Reference:
Muscular energy (from rider) → Kinetic energy (of bicycle) + Heat energy (due to friction)
(Page 125, Exercise Q7)
21Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
Answer & Reference:
No transfer of energy to the rock occurs as there is no displacement. The energy spent by you is converted to heat energy in your muscles and body.
(Page 125, Exercise Q8)
22A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
23An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.
Answer & Reference:
Potential energy at 5 m = mgh = 40 × 10 × 5 = 2000 J
At half-way down (2.5 m), potential energy = 40 × 10 × 2.5 = 1000 J
By conservation of energy, kinetic energy = Total energy - PE = 2000 - 1000 = 1000 J
(Page 125, Exercise Q10)
24What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
Answer & Reference:
Work done by gravity on a satellite is zero. The gravitational force acts towards the center of the earth (centripetal force), while the satellite's displacement is perpendicular to this force (tangential direction). Since force and displacement are perpendicular, work done = F × s × cos(90°) = 0.
(Page 125, Exercise Q11)
25Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.
Answer & Reference:
Yes, according to Newton's first law of motion, an object in uniform motion will continue to move with constant velocity (and thus be displaced) even when no net force acts on it.
(Page 126, Exercise Q12)
26A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.
Answer & Reference:
No work is done on the bundle of hay. Although the person applies an upward force to hold it, there is no displacement of the bundle in the direction of the force. The tiredness is due to muscular effort and energy expenditure in the body, not work done on the hay.
(Page 126, Exercise Q13)
27An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Answer & Reference:
Power = 1500 W = 1.5 kW, Time = 10 h
Energy consumed = Power × Time = 1.5 kW × 10 h = 15 kWh = 15 units
(Page 126, Exercise Q14)
28Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
Answer & Reference:
When the pendulum bob is drawn to one side, it has maximum potential energy and zero kinetic energy. As it swings down, PE converts to KE. At the lowest point, it has maximum KE and minimum PE. As it rises to the other side, KE converts back to PE.
The bob eventually comes to rest due to air resistance and friction at the pivot point, which dissipate energy as heat. The total energy (mechanical + heat) is conserved, so it's not a violation of the law of conservation of energy.
(Page 125-126, Exercise Q15)
29An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?
Answer & Reference:
Work done = Change in kinetic energy = 0 - ½mv² = -½mv²
The negative sign indicates work is done against the motion of the object.
(Page 126, Exercise Q16)
30Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?
Answer & Reference:
Mass (m) = 1500 kg, Velocity (v) = 60 km/h = 60 × (1000/3600) = 50/3 m/s
Work required = Change in KE = 0 - ½mv² = -½ × 1500 × (50/3)²
= -½ × 1500 × 2500/9 = -208333.33 J ≈ -208.33 kJ
(Page 126, Exercise Q17)
31In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.
Answer & Reference:
(Note: Diagrams not provided in source material, but based on standard cases)
• Case I: Force perpendicular to displacement → Work = 0
• Case II: Force in direction of displacement → Work = Positive
• Case III: Force opposite to displacement → Work = Negative
(Page 126, Exercise Q18)
32Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
Answer & Reference:
Yes, I agree with Soni. If multiple forces acting on an object are balanced (net force = 0), then according to Newton's second law (F = ma), acceleration will be zero. For example, a book resting on a table has gravitational force and normal force acting on it, but acceleration is zero.
(Page 126, Exercise Q19)
33Find the energy in joules consumed in 10 hours by four devices of power 500 W each.
Answer & Reference:
Total power = 4 × 500 W = 2000 W = 2 kW
Time = 10 h
Energy consumed = Power × Time = 2 kW × 10 h = 20 kWh = 20 × 3.6 × 10⁶ J = 7.2 × 10⁷ J
(Page 126, Exercise Q20)
34A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
Answer & Reference:
When the object hits the ground, its kinetic energy is converted into other forms of energy: heat, sound, and deformation energy (which may cause the ground or object to deform). The total energy is conserved.
(Page 126, Exercise Q21)