Class 9 Science Chapter 7 Exploration Work Energy and Simple Machines Question Answer Looking for the Class 9 Science Chapter 7 Exploration Work, Energy and Simple Machines Question Answer? You’re in the right place! This chapter explains the concepts of work, energy, power, simple machines, mechanical advantage, and efficiency in a simple and easy-to-understand manner.
Revise, Reflect, Refine (NCERT Textbook Page No. 137)
Question 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.
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).
Question 2.
Fill in the blanks.
(i) Work done = ………………… × ……………………. (in the direction of force).
Answer:
Force, displacement
(ii) 1 joule of work is done when a force of ………………… newton displaces an object by 1 metre in the direction of the force.
Answer:
1
(iii) The expression for kinetic energy of a body of mass m and velocity v is …………… .
Answer:
12mv2
(iv) The potential energy of an object of mass m at a small height h from the Earth’s surface is ………………… .
Answer:
mgh
(v) Power is defined as the ………………….. at which work is done.
Answer:
rate
Question 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.
Answer:
(i) False: Gravity (mg) still acts downward at the highest point.
(ii) False: Acceleration, g = 10 m/s2 (downward) at highest point.
(iii) True: At highest point, velocity O, so KE= 12 mv2 = 0.
(iv) True: All KE has converted to PE. PE is maximised at the highest point.
Question 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 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
Question 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 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).
Question 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.
Answer:
Let height per floor = d, mass = m
10th floor: Height = 10 d. Energy = mg(10d) = 10mgd
20th floor: Height = 20d. Energy = mg(20d) = 2omgd.
Extra energy = 10mgd (double the energy needed).
Power: P1 = 10mgd/t. P2 = 20 mgd/(2t) = 10 mgd/t.
power required is the same (same power, double the energy, double the time).
Question 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.
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.
Question 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.
Answer:
day 1: Mass = 100 kg (scooter) + 60 kg (man) = 160 kg. Day 2:160 kg + 40 kg = 200 kg.
Both days reach same speed v from rest in same time:
KE at Day 1 = 12 × 160 × v2 = 80v2.
KE at Day 2 = 12 × 200 × v2 = 100 v2.
Fuel ratio = 80 v2 : 100 v2 = 4: 5
Question 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.
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.
Question 10.
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:
(i) Sign of work done by gravity: During upward motion, negative (gravity opposes
displacement). During downward motion, positive (gravity aids displacement).
(ii) Height reached: Using v2 = u2– 2gh, 0 = (20)2 – 2 × 10 × h, h= 20m. But actual height is 19.4 m (given), so air resistance did work.
Work done by air resistance = change in total energy. Wair = mghactual – 12 mv2
= 2 × 10 × 19.4 – 12 × 2 × 400 = 388 – 400 = -12 J
Question 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 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
(i) Speed at 0 m:
KE= 12 mv2
180 = 12 × 10 × v2
180 = 5v2 ⇒ v2= 36
v = 6m/s
(ii) Speed at 4 m:
Work = Area under graph
From 0 to 1 m (triangle):
W1 = 12 × 1 × 50 = 25 J
From 1 to 3 m (rectangle):
W2 = 2 × 50= 100 J
From 3 to 4 m (triangle):
W2 = 12 × 1 × 50 = 25 J
Total Work:
W = 25 + 100 + 25 = 150J
Final KE:
KEfinal =180 + 150 = 330J
Final Speed:
12 mv2 = 330
5 v2= 330 ⇒ v2 = 66
v = 66−−√ ≈ 8.1 m/s
Question 12.
The gravitational attraction on the surface of the Moon (lunar surface) is about 16 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:
On Earth: Ball thrown up to height h = 8 m with velocity u.
Using v2 = u2 – 2gearth × h,
u2 = 2 × 10 × 8 = 160 m2 S-2
On Moon: gmoon = gearth 6=106= ms-2 ≈ 1.67 ms-2
Height on Moon: hmoon = u2(2×gmoon )=160(2×106)
160 × 620 =48m
The ball will travel up to 48 m on the Moon with the same initial upward velocity.
Question 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?
Answer:
(i) Between A and B: The car moves at constant speed (35 ms-1) – uniform motion, no change in speed.
(ii) Kinetic energy at A: KE = 12 mv2 = 12 × 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.
Question 14.
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 = (2×10/0.5)−−−−−−−−−−√=40−−√ ≈ 6.32 ms-1
2. 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 = (2×20/0.5)−−−−−−−−−−√ = 80−−√ ≈ 8.94 ms-1.
Question 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 ms-2
Answer:
Mass = 1.5 kg, height = 10 m, g = 10 ms-2.
(i) Velocity just before hitting the sand:
Using conservation of energy: 12 mv2 = mgh, v = 2gh)−−−−√=(2×10×10)−−−−−−−−−−−√ = 200−−−√ ≈ 1.41 ms-1
(ii) Depth of depression: All energy used to create depression. Resistive force of sand = 3000 N. Work done by sand = KE of coconut just before impact = mgh
= 1.5 × 1o × 10 = 150 J
⇒ 3000 N × Depth = 150J
Depth = 150/3000 = 0.05 m = 5 cm
Class 9 Science Chapter 7 Work Energy and Simple Machines Question Answer (InText)
Think It Over (NCERT Textbook Page No. 116)
Question 1.
What will be the magnitude of velocity of the child at the bottom of the blue slide?
Answer:
Using conservation of energy: 12 mv2 = mgh, so v = 2gh−−−√.
The velocity depends only on the height h of the slide, not on the mass of the child or the shape of the slide (neglecting friction).
Question 2.
Will two children of different masses reach the bottom of the same slide with the same velocity?
Answer:
Yes. Since v = 2gh−−−√, 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.
Question 3.
Which slide will result in the largest magnitude of velocity at the bottom?
Answer:
The slide with the greatest vertical height h will give the largest velocity at the bottom, since v = (2gh−−−−√.
Pause and Ponder (NCERT Textbook Page No. 119)
Question 1.
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?
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.
Question 2.
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. Friction acts opposite to the direction of motion (displacement). Since force and displacement are in opposite directions, work done by friction is negative.
Pause and Ponder (NCERT Textbook Page No. 121)
Question 3.
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:
- Kinetic energy of the bicycle and rider,
- Thermal energy (heat) due to friction in wheels, chain, and air resistance,
- Sound energy (minor, from mechanical parts).
Pause and Ponder (NCERT Textbook Page No. 123)
Question 4.
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: 12 mv2A = 12 (4m) v2B, v2A = 4v2B, vAvB=21 Ratio of velocities vA: vB 2: 1.
Question 5.
Does the kinetic energy of an object moving with constant velocity change with its position?
Answer:
No. Kinetic energy KE = 12 mv2. If velocity is constant, KE does not change with position. KE depends only on speed, not on where the object is located.
Pause and Ponder (NCERT Textbook Page No. 126)
Question 6.
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 is the object gradually raised in the vertical direction?
Answer:
Horizontal motion (constant velocity): No change in height h, so U = mgh remains constant. PE does not change.
Vertical motion (raised gradually): Height h increases, so U = mgh increases. PE increases proportionally with height.
Pause and Ponder (NCERT Textbook Page No. 129)
Question 7.
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 = 12 mv2 = 12 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).
Question 8.
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.
Pause and Ponder (NCERT Textbook Page No. 132)
Question 9.
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 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.
Question 10.
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.
Pause and Ponder (NCERT Textbook Page No. 135)
Question 11.
Why is it easier to open the lid of a can by using a spoon as shown in Fig. 7.35?
Answer:
The spoon acts as a Class I lever. The rim of the can acts as the fulcrum. Applying effort at the long end of the spoon creates a larger effort arm. Since MA = effort arm/load arm> 1, a small force at the end of the spoon exerts a much larger force to pop the lid off.
Question 12.
Why do you push an object closer to scissors (fulcrum) when you want to cut an object which is hard?
Answer:
Scissors are a Class I lever. The load arm = distance of object from fulcrum. Moving the object closer to the fulcrum reduces the load arm, which increases MA = effort arm/load arm, so a greater cutting force is exerted on the object for the same effort applied.
Question 13.
Throughout history, many design 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:
All real machines slow down and stop because of energy losses. When a machine works, it uses its energy to do work. At the same time, some energy is always lost due to friction, air resistance, etc. This lost energy is mainly converted into heat (and sometimes sound), which cannot be fully reused.
So, the machine’s total usable energy keeps decreasing. Eventually, no energy is left to do work, and the machine stops. That’s why perpetual motion machines are impossible-they would require no energy loss, which never happens in real life.
Class 9 Science Chapter 7 Question Answer (Activities)
Activity 7.1:
Let Us Investigate (NCERT Textbook Page No. 125)
Aim: To study the relationship between the height from which a ball is dropped and the depth of depression it creates in sand.
Observations:
- When the ball is dropped from a height of 1 m, it creates a depression in the sand.
- When dropped from 2 m, the depression is deeper than from 1 m.
- The depression is deepest when the ball is dropped from the greatest height.
Conclusion:
Raising the ball to a greater height requires more work. Thus, the ball possesses more energy (greater gravitational PE = mgh) at a greater is height.
When released, this PE converts to KE, creating a deeper depression. The greater the height above Earth’s surface, the greater is the potential energy.
Activity 7.2:
Let Us Experiment (NCERT Textbook Page No. 127)
Aim: To demonstrate conservation of mechanical energy using a simple pendulum.
Observations:
Point P (extreme): Pendulum bob is at its highest position. PE = mgh, KF = 0
Point Q (bottom): Pendulum bob is at its lowest position. Potential energy = 0, Kinetic energy is maximum.
Point R (other extreme): Pendulum bob regains its potential energy; KE = 0. Bob almost reaches the same height it started with.
Conclusion:
Mechanical energy (KE + PE) remains constant throughout the oscillation, demonstrating conservation of mechanical energy. In real life, the pendulum gradually slows due to friction at the support and air resistance, which convert mechanical energy to heat.
Activity 7.3:
Let Us Experiment (NCERT Textbook Page No. 131)
Aim: To determine if an inclined plane reduces the force needed to raise an object.
Observation:
The force required to pull the cart up the inclined plank (spring balance reading) is smaller than lifting it vertically.
As the plank becomes less steep (shallower angle), the force required decreases further. However, the distance over which the force is applied increases.
Conclusion:
An inclined plane reduces the effort needed to raise an object to a height, but the effort must be applied over a larger distance. Total work done remains the same (conservation of energy).
Activity 7.4:
Let Us Investigate (NCERT Textbook Page No. 133)
Aim: To demonstrate how a lever can lift a heavier object with a lighter force.
Observations:
Placing the heavier stapler close to the fulcrum (pencil) and a lighter eraser at the far end can lift the stapler.
A much heavier object can be lifted with a lighter effort when the effort arm is larger than the load arm.
Conclusion:
- The lever works on the principle: effort × effort arm = load × load arm.
- By increasing the effort arm, less effort is needed to overcome a large load.
- MA = effort arm/load arm.
Activity 7.5:
Let Us Experiment (NCERT Textbook Page No. 133)
Aim: To verify the law of levers using a beam balance with coins.
Observation:
Result:
Conclusion:
By analysing the values recoded in Table 7.1. The beam balances when n1 × L1 = n1 × L1 (effort effort arm = load × load arm). If the effort arm is increased, the effort required to move the same load is reduced.MA = load/effort = effort arm/load arm. Hence, by increasing the effort arm, the lever applies a larger force F2 to the load than the effort F,. The lever thus allows us to gain a mechanical advantage equal to the ratio of the distances, i.e., L1 / L2.
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