Exploration | Class 9 Science | Chapter 7 - Work, Energy, and Simple Machines | Q&A
CHAPTER 7

Work, Energy, and Simple Machines

Questions & Solutions | Exploration | Class 9 Science

Think It Over Page 136

Think It Over
What will be the magnitude of velocity of the child at the bottom of the blue slide? Will two children of different masses reach the bottom of the same slide with the same velocity? Which of the slides will result in the largest magnitude of velocity for the child at its bottom?
Solution
The velocity at the bottom depends only on the height (v = √(2gh)), not on mass. Both children will have the same velocity. The tallest slide (greatest height) will give the largest velocity.

Ready to Go Beyond

Page 138
The SI unit of work done is joule which is represented by J. The SI unit of force is the newton (N) and the SI unit of displacement is the metre (m). Thus, 1 joule can be defined as 1 J = 1 N × 1 m. That is, 1 joule of work is done on an object when a constant force of 1 newton is applied to it and it is displaced by 1 metre in the direction of the force. Since 1 N = 1 kg m s⁻², note that 1 J = 1 kg m² s⁻². In the graph shown in Fig. 7.4, the force on an object is plotted on the Y-axis against the displacement in the direction of force on the X-axis. In this case, the work done on the object by the force is equal to the area of the shaded rectangle in the graph which is 10 N × 1 m = 10 J. Even when the force is not constant, work done can still be calculated by finding the area under the force-displacement graph between the initial and the final positions.
Solution
Work done = area under force-displacement graph. For constant force, work = F × s.
Page 140
Doing mechanical work is one way of transferring energy from one object to another. But that is not the only way! Energy can also be transferred as heat. When two objects at different temperatures come in contact, energy flows from the hotter one to the colder one. Energy can also move without direct contact. For example, the Sun's energy reaches the Earth through radiation. Energy is transferred in electric circuits, as well as via sound waves, and even in nuclear reactions that power the Sun.
Solution
Energy can be transferred through work, heat, radiation, electricity, sound, and nuclear reactions.
Page 144
You need to apply an external force to overcome the internal forces in the spring to deform it. Once you remove this external force, the internal forces undo the deformation, and in the process, it can carry out work. Thus, internal forces allow energy to be stored in a deformed object.
Solution
Energy is stored in deformed objects as potential energy due to internal forces (e.g., elastic potential energy in springs).
Page 146
Work done on a system against its internal forces, such as gravitational, electric or magnetic forces, can result in a gain of the potential energy of the system. But this is not true for all internal forces. For example, work done against friction does not lead to a storage of energy. You will learn how to identify such forces in higher grades.
Solution
Work against conservative forces (gravity, spring, electric, magnetic) stores potential energy. Work against non-conservative forces (friction) dissipates energy as heat.
Page 147
Mechanical energy is just one part of a bigger picture. In nature, energy can appear in many different forms. Scientists have discovered that the total energy of an object or system of objects which is not acted upon by any external forces, stays constant.
Solution
Law of conservation of energy: total energy in an isolated system remains constant; energy changes from one form to another but is never created or destroyed.
Page 150
Movable pulleys or a system of pulleys (Fig. 7.25) can have a mechanical advantage greater than 1 and can lift much heavier objects with much smaller effort. In a movable pulley system, the load is attached to the movable pulley. One end of the rope is fixed to a point, while the other end is free to apply effort. Pulleys are widely used in real life, such as in elevators and cranes given the convenience they provide us.
Solution
Movable pulleys and pulley systems increase mechanical advantage, reducing the effort needed to lift heavy loads.
Page 155
Levers can be of three classes depending upon the relative positions of effort, fulcrum and load, as shown in Table 7.2. Class I: Fulcrum in between (e.g., seesaw, scissors). Class II: Load in between (e.g., wheelbarrow, bottle opener). Class III: Effort in between (e.g., tweezers, broom).
Solution
Class I lever: fulcrum between effort and load. Class II lever: load between fulcrum and effort. Class III lever: effort between fulcrum and load.

Threads of Curiosity

Page 149
You may have heard about another unit called horsepower (hp) used to measure power, especially for car engines, or pumps used to lift water. One horsepower is equal to 746 W. In the early days, when engines were newly discovered, the powers of engines were compared to the power of actual horses which were used to drive carriages.
Solution
1 horsepower (hp) = 746 watts. It was originally used to compare engine output with the power of draft horses.
Page 151
In everyday life, we say that a vehicle is accelerating when the magnitude of its velocity is changing but we often fail to recognise that there can be acceleration when there is only a change in the direction of velocity.
Solution
Acceleration occurs when either magnitude or direction of velocity changes. Uniform circular motion has constant speed but changing direction, hence acceleration.

Meet a Scientist Page 150

Meet a Scientist - James Watt
The unit of power, watt is named in the honour of James Watt. He invented an efficient steam engine that could generate rotational motion and move wheels.
Solution
James Watt improved the steam engine, making it more efficient, and the unit of power (watt) is named after him.

Bridging Science and Society

Page 140
Microphones that help us capture sound from various sources convert sound energy to electrical energy. When we speak or sing into a microphone, the sound waves make a thin membrane, called a diaphragm, vibrate. These vibrations are converted into an electrical signal. A speaker does the opposite; an electrical signal makes a cone or diaphragm attached inside the speaker vibrate, which produces sound. If all components work properly, the sound from the speaker closely matches the originally captured sound.
Solution
Microphones convert sound energy to electrical energy; speakers convert electrical energy to sound energy.
Page 145
When brakes are applied to a moving vehicle, it moves some distance before coming to a stop. The distance travelled depends upon the velocity of the vehicle when the brakes are applied, the road surface (wet/dry, etc.), the braking capacity of the vehicle (the negative acceleration caused by the brakes) as well as the driver's reaction time. Can you now understand why it is important to maintain a safe distance from the vehicle moving ahead of your vehicle and how this distance needs to be adjusted given your initial velocity? There is a vehicle-to-vehicle (V2V) communication technology, now being developed in many countries including India, which allows vehicles to exchange signals and warns drivers of possible collisions.
Solution
Stopping distance increases with speed. Safe following distance must be adjusted based on speed to allow enough time to stop.
Page 147
When brakes are applied to a moving vehicle, it moves some distance before coming to a stop. The distance travelled depends upon the velocity of the vehicle when the brakes are applied, the road surface, the braking capacity, and the driver's reaction time.
Solution
Braking distance = v²/(2a). Higher speed means longer stopping distance. Reaction time also adds to total stopping distance.

Activity 7.1 Page 145

Activity 7.1
Let us investigate: 1. Take a heavy ball and a large container filled with loose sand. 2. Raise the ball over the sand bed to a height of about 1 m and drop it. Is a depression created in the sand? Why does the ball create a depression? 3. Now, raise the ball to the height of 2 m and release it at a slightly different position. Repeat this step one more time. Compare the depths of the depressions. Is there any difference? In which case is the depression deepest and in which case the shallowest?
Solution
The ball creates a depression because it possesses potential energy at height, which converts to kinetic energy on falling. The deeper the depression, the more energy the ball had. The ball dropped from 2 m creates the deepest depression (more potential energy = mgh).

Activity 7.2 Page 147

Activity 7.2
Let us experiment: 1. Set up a simple pendulum as you have learnt in Grade 7. 2. Paste a white sheet of paper on a wall behind the pendulum. Draw a horizontal line above the position of the bob when it is not oscillating (Fig. 7.20). 3. Take the bob to one side to a point P, which is at the level of the horizontal line and let it go. Observe it at the extreme points of the first couple of oscillations. Does the bob almost reach the level of the horizontal line?
Solution
The bob rises to nearly the same height on the other side, showing conservation of mechanical energy (potential energy converts to kinetic and back to potential). In real life, air resistance and friction cause gradual energy loss.

Activity 7.3 Page 151

Activity 7.3
Let us experiment: 1. Take a smooth plank about 1.5 m long, a toy car, and a spring balance. Arrange an elevated surface at about 0.5 m height. 2. First, lift the cart vertically and note the spring balance reading (weight). 3. Next, place the plank against the top and pull the cart along the plank. Is the reading smaller than that of step 2? 4. Now, reduce the angle between the plank and the base and repeat step 3. Observe how the force required changes as the plank becomes less steep.
Solution
The force required decreases as the incline becomes less steep (longer plank). Work done remains the same (mgh), but force decreases as distance increases.

Activities 7.4 & 7.5 Page 153

Activity 7.4
Let us investigate: 1. Take a 30 cm long scale, a pencil, 2-3 erasers and a stapler. 2. Place the scale over the pencil such that the pencil is closer to one end. On the end of the scale closer to the pencil, place the stapler. 3. On the other end, place one eraser. Does the stapler lift up? If not, add one more eraser. You may have noticed that a much heavier object could be lifted by a much lighter eraser. This was made possible by using a scale as a simple machine called a lever.
Solution
A lever provides mechanical advantage. By increasing the effort arm distance, a small effort can lift a larger load.
Activity 7.5
Let us experiment: 1. Take a long scale, a piece of string, two paper cups, adhesive tape, and identical coins. 2. Tie the string tightly around the scale at its midpoint. Hang the scale from this string. 3. Fix paper cups to both ends. Check whether the beam is levelled. 4. Place 1 coin in the left pan and 1 coin in the right pan. The beam stays horizontal. 5. Add one more coin to the right pan. The beam tilts. Move the heavier pan closer to the centre to balance the beam. Measure its distance from the centre. 6. Repeat with 4 coins and then 8 coins. 7. Record all observations. You can conclude that the beam balances when n₁ × L₁ = n₂ × L₂.
Solution
Beam balance works on the principle of moments: effort × effort arm = load × load arm.

Pause and Ponder

Page 139 (Q1)
In the previous chapter, a weightlifter is shown holding a barbell steady in her hands (Fig. 6.8). Is she doing any work on the barbell while holding it steady?
Solution
No, because displacement is zero. Work done = force × displacement. No displacement means no work.
Page 139 (Q2)
Is the work done by friction on the stack of coins that travels on a rough surface (Fig. 6.13c) — positive, negative or zero?
Solution
Negative work, because friction acts opposite to the direction of displacement.
Page 140 (Q3)
When you pedal a bicycle on a flat road, your muscles supply energy. In what forms does this muscular energy appear as you ride?
Solution
Muscular energy converts to kinetic energy (motion of bicycle), heat energy (friction), and sound energy.
Page 143 (Q4)
Two objects A and B of mass m and 4m have the same kinetic energy. What is the ratio of the magnitude of velocities of A and B?
Solution
KE = ½mv². If KE same, then ½m v_A² = ½(4m) v_B² → v_A² = 4v_B² → v_A/v_B = 2:1.
Page 143 (Q5)
Does the kinetic energy of an object which moves with constant velocity change with its position?
Solution
No, kinetic energy depends only on mass and speed, not on position. Constant velocity means constant kinetic energy.
Page 145 (Q6)
Does the potential energy of an object near the surface of the Earth change if it moves with constant velocity in the horizontal direction? What if the object is gradually raised in the vertical direction?
Solution
Horizontal motion: potential energy does not change (height constant). Vertical motion: potential energy increases with height (U = mgh).
Page 146 (Q7)
For the situation depicted in Fig. 7.19, calculate the mechanical energy of the ball just before it hits the ground and show that even at this position, it is mgh.
Solution
At ground, potential energy = 0, kinetic energy = ½mv² = ½m(2gh) = mgh. Total mechanical energy = mgh (conserved).
Page 146 (Q8)
You may have seen an exhibit like that in Fig. 7.22 in a science park, where a ball is released from the highest point. Describe how the kinetic energy and potential energy change at points A, B and C. Why do subsequent points, such as C, D and E, usually have lower heights compared to the previous ones? Could it have anything to do with the energy lost due to friction?
Solution
At highest point: max PE, min KE. At lowest point: min PE, max KE. Lower heights after each cycle are due to energy loss from friction and air resistance.
Page 149 (Q9)
Explain why roads on hills are built to wind around in gentle slopes rather than going straight up (Fig. 4.26)?
Solution
Winding roads increase the length of the incline, reducing the force needed to climb (mechanical advantage of inclined plane).
Page 149 (Q10)
To reach a higher floor, we find climbing an inclined ladder easier in comparison to climbing a vertical ladder (Fig. 7.30). Explain why.
Solution
An inclined ladder reduces the force required (mechanical advantage) because the same height is achieved over a longer distance.
Page 152 (Q11)
Why is it easier to open the lid of a can by using a spoon as shown in Fig. 7.35?
Solution
The spoon acts as a lever (Class I), increasing the force applied to the lid (mechanical advantage).
Page 152 (Q12)
Why do you push an object closer to scissors fulcrum when you want to cut an object which is hard?
Solution
Placing the object closer to the fulcrum reduces the load arm, increasing the mechanical advantage of the scissors.
Page 152 (Q13)
Throughout history, many designs of perpetual machines (using wheels, weights or magnets) have been proposed but none actually work. Why do all real machines eventually slow down and stop? Explain in terms of work and energy.
Solution
Energy is lost to friction, air resistance, and heat. No machine is 100% efficient; some energy is always dissipated, causing the machine to eventually stop.

What if... Page 155

What if...
it were possible to build a perpetual motion machine, which once started, could continue doing useful work forever, without any fuel or electricity?
Solution
Perpetual motion machines violate the law of conservation of energy. Energy cannot be created, and some energy is always lost to friction and heat, so they are impossible.

END-OF-CHAPTER EXERCISES Pages 156-158

1.
State whether True or False. (i) Work is said to be done when a force is applied, even if the object does not move. (ii) Lifting a bucket vertically upward results in positive work done on the bucket. (iii) The SI unit for both work and energy is joule (J). (iv) A motionless stretched rubber band has kinetic energy. (v) Energy can change from one form to another.
Solution
(i) False (displacement required), (ii) True, (iii) True, (iv) False (it has potential energy), (v) True.
2.
Fill in the blanks. (i) Work done = ______ × ______ (in the direction of force). (ii) 1 joule of work is done when a force of ______ newton displaces an object by 1 metre in the direction of the force. (iii) The expression for kinetic energy of a body of mass m and velocity v is ______. (iv) The potential energy of an object of mass m at a small height h from the Earth's surface is ______. (v) Power is defined as the ______ at which work is done.
Solution
(i) force × displacement, (ii) 1, (iii) ½mv², (iv) mgh, (v) rate.
3.
When a ball thrown upwards reaches its highest point, tick which of the following statement(s) are correct? (i) The force acting on the ball is zero. (ii) The acceleration of the ball is zero. (iii) Its kinetic energy is zero. (iv) Its potential energy is maximum.
Solution
Options (iii) and (iv) are correct. At highest point, velocity = 0 so KE = 0, and height is maximum so PE = maximum.
4.
For each of the following situations, identify the energy transformation that takes place: (i) a truck moving uphill, (ii) unwinding of a watch spring, (iii) photosynthesis in green leaves, (iv) water flowing from a dam, (v) burning of a matchstick, (vi) explosion of a fire cracker, (vii) speaking into a microphone, (viii) a glowing electric bulb, and (ix) a solar panel.
Solution
(i) Chemical → Kinetic + Potential, (ii) Elastic potential → Kinetic, (iii) Light → Chemical, (iv) Potential → Kinetic, (v) Chemical → Heat + Light, (vi) Chemical → Heat + Light + Sound, (vii) Sound → Electrical, (viii) Electrical → Light + Heat, (ix) Light → Electrical.
5.
A student is slowly lifted straight up in an elevator from the ground level to the top floor of a building. Later, the same student climbs the staircase, all the way to the top. Given that the height of the building is h = 72.5 m, acceleration due to gravity is g = 10 m/s², and student's mass is m = 50 kg. (i) Find the gain in the potential energy if the student is lifted straight up to the top. (ii) Find the gain in the potential energy when the student climbs the stairs to the same top. (iii) What do you conclude about the dependence of the potential energy on the path taken?
Solution
(i) PE = mgh = 50 × 10 × 72.5 = 36,250 J. (ii) Same = 36,250 J. (iii) Potential energy depends only on height, not on the path taken (conservative force).
6.
A crane lifts a mass m to the 10th floor of a building in a certain time. It then raises the same mass to the 20th floor of the same building in double the time. How much more energy and power are required? Assume that the height of all floors is equal.
Solution
Energy is proportional to height (mgh). 20th floor requires double the energy of 10th floor. Power = energy/time. Power remains the same because energy doubles and time doubles.
7.
Which factors determine the energy required to raise a flag from the ground to the top of a tall flagpole using a pulley? Does raising the flag slowly or quickly change the amount of work done? If the speed at which the flag is raised is doubled, how does the power requirement change? Explain your answers.
Solution
Energy required = mgh (mass, gravity, height). Work done is the same regardless of speed. Power = work/time, so doubling speed doubles power (since time halves).
8.
A man of mass 60 kg rides a scooter of mass 100 kg. He accelerates the scooter to a velocity v. The next day, his son with a mass of 40 kg joins him as a passenger. If the scooter reaches the same speed on both days in the same time interval, what is the ratio of the fuel of the tank used on the two days? Assume that the energy transfer to the scooter happens entirely due to fuel, and no other losses occur due to air resistance and friction.
Solution
KE = ½mv². Day 1 mass = 160 kg, Day 2 mass = 200 kg. Ratio of KE = 160:200 = 4:5. Fuel used is proportional to KE, so ratio = 4:5.
9.
On a seesaw with sliding seats, a child is sitting on one side and an adult on the other side. The adult weighs twice that of the child. The seesaw however is balanced. Draw a figure which depicts this situation showing the distances from the fulcrum where the child and the adult are seated.
Solution
For balance: child's weight × child's distance = adult's weight × adult's distance. Since adult weighs twice, adult sits at half the distance from fulcrum compared to child.
10.
A ball of mass 2 kg is thrown up with a velocity of 20 m/s. (i) Identify the sign of the work done by gravity on the ball during its upward motion and its downward motion. (ii) If the ball reaches a height of 19.4 m, how much work was done by air resistance (assume g = 10 m/s²).
Solution
(i) Upward: negative work by gravity; Downward: positive work by gravity. (ii) Without air resistance, height = u²/2g = 400/20 = 20 m. Actual height = 19.4 m. Work by air resistance = loss in mechanical energy = mg(20 - 19.4) = 2×10×0.6 = 12 J (negative work).
11.
A 10.0 kg block is moving on horizontal floor with negligible friction. As shown in the Fig. 7.37, a variable force is applied on the block in its direction of motion from its position at 0 m till 4 m. If the block had a kinetic energy of 180 J when it was at 0 m, find the block's speed (i) at 0 m, and (ii) at 4 m. Does the block have negative acceleration in any portion of its motion?
Solution
(i) KE = ½mv² → 180 = ½×10×v² → v² = 36 → v = 6 m/s at 0 m. (ii) Work done = area under F-s graph. Add to initial KE to find final KE, then v = √(2KE/m). Negative acceleration occurs where force is opposite to motion (negative F).
12.
The gravitational attraction on the surface of the Moon (lunar surface) is about 1/6 th of that on the surface of the Earth. An astronaut can throw a ball up to a height of 8 m from the surface of the Earth. How far up will the ball thrown with the same upward velocity travel from the surface of the Moon?
Solution
On Earth: v² = 2gh (g = 10 m/s², h = 8 m). On Moon: g' = g/6 = 10/6 m/s². Same v, so h' = v²/(2g') = (2gh)/(2g/6) = 6h = 48 m.
13.
A 1000 kg car is moving along a road at a constant speed. Suddenly, the driver notices some obstruction ahead and applies the brakes to come to a complete stop. The graphical representation of motion of the car starting from the instant the driver spots the traffic ahead is shown in Fig. 7.38. (i) Describe how the car moves between positions A and B. (ii) Calculate the kinetic energy of the car at A. (iii) State the work done by the brakes in bringing the car to a halt between B and C. (iv) What does the kinetic energy of the car transform into?
Solution
(i) Constant velocity between A and B (reaction time). (ii) KE = ½ × 1000 × (velocity from graph)². (iii) Work by brakes = -KE at A (negative work). (iv) Kinetic energy transforms into heat energy (brakes heat up) and sound energy.
14.
The potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown in Fig. 7.39. At O, the velocity of the ball is 0 m/s and potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.
Solution
Total mechanical energy = PE + KE = constant = 30 J at O (since KE = 0). At each point, KE = Total - PE, then v = √(2KE/m). Calculate using graph values.
15.
A coconut of mass 1.5 kg falls from the top of a coconut tree onto the wet sand on a beach. The height of the tree is 10 m. On impact, the coconut comes to rest by making a depression in the sand. (i) Calculate the velocity of the coconut just before it hits the sand. (ii) Assume that the average resistive force of sand is 3000 N and all of the coconut's energy is used to create the depression in the sand. Calculate the depth of the depression the coconut makes in the sand. Assume g = 10 m/s².
Solution
(i) v = √(2gh) = √(2×10×10) = √200 = 10√2 ≈ 14.14 m/s. (ii) KE just before impact = mgh = 1.5×10×10 = 150 J. Work by sand = F × d = 3000 × d = 150 J → d = 150/3000 = 0.05 m = 5 cm.

Chapter Summary

  • Work done = force × displacement in the direction of force. Unit: joule (J).
  • Positive work: force and displacement in same direction. Negative work: force opposite to displacement. Zero work: no displacement or force perpendicular to displacement.
  • Kinetic energy (KE) = ½mv². Potential energy (PE) = mgh (gravitational).
  • Work-energy theorem: Work done = change in energy.
  • Conservation of mechanical energy: KE + PE = constant (in absence of non-conservative forces).
  • Power = work/time = energy/time. Unit: watt (W). 1 W = 1 J/s.
  • Simple machines: pulley, inclined plane, lever. Mechanical advantage = load/effort.
  • Machines do not create energy; they only transfer or transform it.

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📚 Exploration | Class 9 Science | NCERT
Chapter 7: Work, Energy, and Simple Machines — All Questions with Solutions

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