NCERT Solutions for Class 9 science Chapter 7: Work Energy and Simple Machines

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.
Answer:
(i) False: Work requires both force and displacement in the direction of force.
(ii) True: Lifting a bucket upward: force and displacement both upward, positive work.
(iii) True: SI unit of both work and energy is joule (J).
(iv) False: A motionless stretched rubber band has potential energy (elastic PE),
not kinetic energy.
(v) True: Energy can change from one form to another (law of conservation of energy).

Fill in the blanks.

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.
Answer:
(i) The force acting on the ball is zero.
False: Even at the highest point, gravity acts downward on the ball. The force is not zero.

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 firecracker,
(vii) speaking into a microphone,
(viii) a glowing electric bulb, and
(ix) a solar panel.
Answer:
(i) Truck uphill: KE to PE (kinetic energy converts to gravitational potential energy)
(ii) Unwinding watch spring: Elastic PE to KE (mechanical energy of spring to motion of hands)
(iii) Photosynthesis In green leaves: Light energy to Chemical energy
(iv) Water flowing from dam: PE to KE (potential to kinetic), then KE to Electrical energy
(v) Burning matchstick: Chemical energy to Thermal energy + Light energy.
(vi) Explosion of firecracker: Chemical energy to KE + Sound energy + Light energy + Heat energy
(vii) Speaking into microphone: Sound energy to Electrical energy
(viii) Glowing electric bulb: Electrical energy to Light energy + Thermal energy (heat)
(ix) Solar panel: Light energy (solar) to Electrical energy

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 ms-2, 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?
Answer:
(i) PE gained (elevator): U = mgh = 50 × o × 72.5 = 36,250 J
(ii) PE gained (staircase): Same = 36,250 J (height is same)
(iii) Conclusion: Potential energy depends only on the height h gained and the mass m, not on the path taken (elevator or staircase).

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.
Answer:
Crane lifting a mass

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.
Answer:
Energy required: Determined by mass of flag (m), height of flagpole (h), and g. E = mgh. Speed of raising: Does not change the work done (W = mgh regardless of speed). But power changes: P = W/t. Raising quickly (less time) requires more power. If the speed is doubled, the time taken is reduced to half, but the power required becomes twice as much.

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.
Answer:
The fuel used is proportional to the kinetic energy imparted to the system.
m1 (Day 1) = 100 kg + 60 kg = 160 kg
m2 (Day 2) = 160 kg + 40 kg = 200 kg
Ratio of fuel =
=
=
=
The ratio of fuel used is 4 : 5.

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.
Answer:
The adult weighs twice that of the child. The seesaw is balanced. Let mass of child = m, then mass of adult = 2m. For balance: effort × effort arm = load × load arm, so m × dChild = 2m × dadult, meaning d child = 2 x dadult.
The child must sit at twice the distance from the fulcrum compared to the adult.

A ball of mass 2 kg is thrown up with a velocity of 20 ms-1.
(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 ms-2)
Answer:
Ball thrown upward (mass = 2 kg, velocity = 20 m/s)
(i) Upward motion:
Gravity acts downward, displacement is upward → work done by gravity is negative.
Downward motion:
Gravity acts downward, displacement is downward → work done by gravity is positive.

A 10.0 kg block is moving on a horizontal floor with negligible friction. As shown in Fig. 7.37, a variable force is applied to the block in its direction of motion from its position at o 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?

Answer:
Given,
Mass m = 10 kg
Initial KE= 10 J

The gravitational attraction on the surface of the Moon (lunar surface) is about 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?
Answer:
Kinetic energy is converted to potential energy.
½ mv2 = mgehe (On Earth)
½ mv2 = mgmhm (On Moon)
Equating the potential energies,
gee = gmhm
ge (8 m) = ≤ft() hm
hm = 8 m × 6 = 48 m

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?
Answer:
(i) Between A and B: The car moves at a constant speed (35 ms-1) – uniform motion, no change in speed.
(ii) Kinetic energy at A: KE = mv2 = × 1000 × (35)2 = 612,500 J = 6.125 × 105 J
(iii) Work done by brakes (B to C): The car decelerates from 35 m.s-1 to 0. Work done by brakes = -KE at B = -612,500 J
(iv) KE transforms into: Thermal energy (heat in brake pads and road) and some sound energy.

The potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown in Fig. 7.39. At 0, the velocity of the ball is 0 ms-1 and potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.

Answer:
A 0.5 kg ball moves along a frictionless track.
At 0,v = 0 ms-1 and PE = 30J.
So, total mechanical energy
= KE+ PE = 0 + 30 = 30J.
1. At P(PE = 20J):KE = 30 – 20 = 10 J,
v = ≈ 6.32 ms-1

At Q (PE = 30J) : KE = 30 – 30 = 0 J,
v = 0 ms-1
3. At R (PE = 40J) : If PE = 40J > total E, this point would be inaccessible. From graph, at R, PE = 40 J, KE would be negative, which is impossible, the ball cannot reach R unless it gains energy.

If total E = 30 J, only points where PE ≤ 30 J are accessible. Note: Reading R from figure as PE= 10J,KE = 20J, v = = ≈ 8.94 ms-1.

What will be the magnitude of the velocity of the child at the bottom of the blue slide?
Answer:
When a child slides down, their potential energy at the top is converted into kinetic energy at the bottom (ignoring friction).
At the top: Potential Energy = (mgh)
At the bottom: Kinetic Energy = (½ mv2)
Equating:
mgh = ½ mv2
⇒ v2 = 2 gh
⇒ v =

Will two children of different masses reach the bottom of the same slide with the same velocity?
Answer:
Yes. Since v = , velocity at the bottom depends only on height h and acceleration due to gravity g. Mass cancels out, so both children reach the bottom with the same speed.

Which slide will result in the largest magnitude of velocity at the bottom?
Answer:
The slide that will give the child the largest velocity at the bottom is the one with the greatest vertical height.
At the top of the slide, the child has gravitational potential energy = mgh.
At the bottom, this energy is converted into
kinetic energy = ½ mv2
Equating: ½ mv2 = mgh
We know that the velocity depends only on height (h) and gravity (g), not on the mass of the child or the shape of the slide.

In previous chapter, a weight lifter is shown holding a barbell steady in her hands (Fig.6.8). Is she doing any work on the barbell while holding it steady?
Answer:

No. Since the barbell has zero displacement (s = 0), work done W = F × s = F × 0 = 0 J. The weightlifter feels tired because her muscles keep contracting and expanding, using internal energy, but scientifically, no work is done on the barbell.

Is the work done by friction on the stack of coins that travels on a rough surface (Fig.6. 13e) positive, negative, or zero?

Answer:
Negative; kinetic friction acts in the direction opposite to the relative motion (displacement) of the stack of coins. Because the force and displacement are in opposite directions (an angle of 180°), the work done by friction is negative.

When you pedal a bicycle on a flat road, your muscles supply energy. In what forms does this muscular energy appear as you ride?
Answer:
Muscular energy is converted into:

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’
Answer:
KE is same: mv2A = (4m) v2B, v2A = 4v2B, Ratio of velocities vA: vB 2: 1.

Does the kinetic energy of an object moving with constant velocity change with its position?
Answer:
No. Kinetic energy KE = mv2. If velocity is constant, KE does not change with position. KE depends only on speed, not on where the object is located.

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?
Answer:
No, the potential energy does not change when an object moves horizontally at a constant velocity. This is because its height relative to the ground remains the same, and potential energy (p = mgh) only depends on vertical position. However, if the object is gradually raised in the vertical direction, its potential energy increases. As height increases, the work done against gravity is stored as additional gravitational potential energy.

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.

Answer:
Just before hitting the ground, h’ = 0, so PE = 0. KE = mv2 = m(2gh) = mgh
Therefore, total mechanical energy = KE + PE = mgh + 0 = mgh. This equals the initial mechanical energy at height h (where v = 0, PE = mgh).

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 changes 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?

Answer:
When the ball starts from the top, it has more potential energy and less kinetic energy. At A and C (higher points): more potential energy, less kinetic energy At B (lowest point): less potential energy, more kinetic energy. The heights at C, D and E are lower because some energy is lost due to friction.

Explain why roads on hills are built to wind around in gentle slopes rather than going straight up.
Answer:
A winding road acts as an inclined plane of longer length L and a smaller angle. MA = L/ h, so a longer inclined plane gives greater mechanical advantage, requiring less effort (force) to drive up the hill. The total work done remains the same, but the force required at each point is much smaller.

To reach a higher floor, we find climbing an inclined ladder easier in comparison to climbing a vertical ladder (Fig. 7.30). Explain why.

Answer:
An inclined ladder acts like an inclined plane. The person applies force along the incline over a longer distance (L> h), so the force required at each step is less than their full body weight. MA = L/h> 1. The total work done is the same, but the effort per step is reduced.

Why is it easier to open the lid of a can by using a spoon as shown in Fig. 7.35?

Answer:
The spoon functions as a class I lever. The rim of the can acts as the fulcrum. Because the handle is long (large effort arm) and the tip is short (small load arm), the MA = Effort arm/Load arm is very large. This greatly multiplies the applied force, making it easier to open the lid.

Why do you push an object closer to scissors (fulcrum) when you want to cut a hard object?
Answer:
Scissors are double-levers, with the screw in the middle acting as the fulcrum. The force you apply with your hand is amplified when the item being cut is closer to this fulcrum. Placing the object closer to the pivot point reduces the load arm (distance from the object to the pivot), maximizing the cutting force because less effort is lost to leverage.

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.
Answer:
In terms of work and energy, no real machine can run forever because of friction and air resistance. As the parts of a machine move, they must do work (W = F × s) to overcome these resistive forces. According to the law of conservation of energy, energy cannot be created; it can only be transformed from one form to another.