Work, Energy, and Simple Machines Class 9 — Competency Based Questions

20 Questions | 75 Marks

A farmer is trying to pull a heavy cart filled with sugarcane. He exerts a force on the cart, but it doesn't move. According to the scientific definition of work, has the farmer done any work on the cart? Justify your answer. 3marks

A satellite orbits the Earth in a circular path. A gravitational force acts on the satellite towards the Earth.(a) Is the work done by the gravitational force on the satellite positive, negative, or zero?(b) Explain your reasoning based on the concepts of force and displacement. 2marks

A student pushes a wall with a force of 100 N for 2 minutes, but the wall does not move.(a) Calculate the work done by the student on the wall.(b) Explain why the student might feel tired despite doing no work in the scientific sense. 3marks

A constant force of 20 N is applied to an object, and it moves through a distance of 10 m in the direction of the force.(a) Calculate the work done by the force.(b) If the object has a mass of 5 kg, what is its acceleration if it started from rest? 3marks

Assertion: When a person holds a heavy bag of groceries steady, they do not perform any scientific work on the bag.Reason: Work done is defined as force multiplied by displacement in the direction of the force. Since there is no displacement, no work is done.(a) Both assertion and reason are true, and the reason is the correct explanation of the assertion.(b) Both assertion and reason are true, but the reason is not the correct explanation of the assertion.(c) Assertion is true, but the reason is false.(d) Assertion is false, but the reason is true. 1mark

A body of mass 10 kg is moving with a velocity of 5 m/s. A force is applied to it, and it starts moving with a velocity of 10 m/s.(a) Calculate the initial kinetic energy of the body.(b) Calculate the final kinetic energy of the body.(c) Calculate the work done by the force on the body. 4marks

A car of mass 1200 kg is moving at a speed of 54 km/h. The driver applies brakes and the car stops in 5 seconds.(a) Calculate the initial kinetic energy of the car.(b) Calculate the work done by the brakes.(c) What is the magnitude of the braking force, assuming it to be constant? 5marks

Case Study: A family is on a picnic at the top of a hill. They decide to roll a ball down a gentle slope towards a lake at the bottom. The height of the hill is 30 m. The mass of the ball is 0.5 kg. Assume no energy losses due to friction or air resistance.

(a) What is the potential energy of the ball at the top of the hill? (Take g = 10 m/s²) [1 mark]
(b) What is the kinetic energy of the ball at the top of the hill? [1 mark]
(c) Calculate the velocity of the ball when it reaches the bottom of the hill. [2 marks]
(d) Explain why the velocity of the ball at the bottom of the hill depends only on the height and not on the mass of the ball. [1 mark]

An object of mass 5 kg is raised to a height of 10 m.(a) Calculate the work done against gravity.(b) What is the potential energy gained by the object? 2marks

Two objects, A and B, have the same kinetic energy. Object A has mass

m_{A}

and velocity

v_{A}

, while object B has mass

m_{B}

and velocity

v_{B}

. If

m_{A} = 4 m_{B}

, what is the ratio of their velocities,

v_{A} / v_{B}

? 3marks

Assertion: When a person lifts a book from a table and places it on a higher shelf, the potential energy of the book increases.Reason: Potential energy is stored energy due to the position of an object.(a) Both assertion and reason are true, and the reason is the correct explanation of the assertion.(b) Both assertion and reason are true, but the reason is not the correct explanation of the assertion.(c) Assertion is true, but the reason is false.(d) Assertion is false, but the reason is true. 1mark

A toy car of mass 0.5 kg is moving with a speed of 2 m/s. It then rolls up a ramp and comes to rest at a height of 0.2 m.(a) Calculate the initial kinetic energy of the toy car.(b) Calculate the potential energy gained by the toy car.(c) How much energy is lost due to friction during this process? (Take g = 10 m/s²) 4marks

A water pump lifts 5000 litres of water to a height of 20 m in 1 minute. Calculate the power of the pump. (Assume density of water = 1000 kg/m³ and g = 10 m/s²) 4marks

A weightlifter lifts a mass of 200 kg to a height of 1.5 m in 5 seconds. Calculate the power exerted by the weightlifter. (Take g = 10 m/s²) 3marks

Assertion: A lever can multiply force but not work.Reason: In an ideal lever, the product of effort and effort arm is equal to the product of load and load arm.(a) Both assertion and reason are true, and the reason is the correct explanation of the assertion.(b) Both assertion and reason are true, but the reason is not the correct explanation of the assertion.(c) Assertion is true, but the reason is false.(d) Assertion is false, but the reason is true. 1mark

A girl uses a pulley system to lift a box of mass 10 kg to a height of 5 m. If the effort applied is 50 N, calculate:(a) The work done against gravity (load).(b) The work done by the effort.(c) The mechanical advantage of the pulley system.(d) Is this a fixed pulley, movable pulley, or a block and tackle system? Justify. (Take g = 10 m/s²) 5marks

Case Study: Consider a child playing on a swing. The child moves back and forth, reaching different heights at the extreme points. The swing is attached to two fixed points by ropes.

(a) At the highest point of the swing, what form of energy is predominantly present? [1 mark]
(b) When the swing is at its lowest point, what form of energy is predominantly present? [1 mark]
(c) If we ignore friction and air resistance, what principle describes the transformation of energy between the highest and lowest points? [1 mark]
(d) Explain why the swing eventually slows down and stops in reality. [1 mark]

A family pushes a car of mass 1000 kg to move it up an inclined road of length 100 m and vertical height 10 m. They exert a force of 500 N along the incline.(a) Calculate the work done by the family.(b) Calculate the potential energy gained by the car.(c) If the car was initially at rest, calculate the kinetic energy of the car at the top of the incline. 4marks

A horizontal force of 50 N is applied to a block of mass 20 kg. The block moves a distance of 10 m.(a) Calculate the work done by the applied force.(b) If there is a frictional force of 20 N acting opposite to the motion, calculate the work done by friction.(c) Calculate the net work done on the block.(d) What is the change in kinetic energy of the block? 4marks

Scenario: In a hydroelectric power plant, water stored at a high potential energy level is released. As it flows down, its potential energy is converted into kinetic energy, which then drives turbines to generate electricity.(a) Explain the energy transformation that occurs from the stored water to the running turbine. [2 marks](b) If the water falls from a height of 50 m and has a mass of 2000 kg, calculate the potential energy it has at the top. (Take g = 10 m/s²) [2 marks](c) If all this potential energy is converted into kinetic energy when it reaches the turbine, what is its velocity just before hitting the turbine? 1mark

Answers

No, the farmer has not done any work on the cart according to the scientific definition. Work is done only when a force causes displacement. Since the cart did not move, the displacement is zero, and hence, the work done is zero, irrespective of the force applied.

(a) Zero.(b) The gravitational force acts radially towards the Earth, while the displacement of the satellite in its circular orbit is tangential to the orbit at every point. Since the force and displacement are perpendicular to each other, the work done is zero.

(a) Work done = Force × Displacement. Since the wall does not move, the displacement is 0 m. Therefore, Work done = 100 N × 0 m = 0 J.(b) The student feels tired because their muscles are exerting a force and using internal energy, even though no external work is done on the wall.

(a) Work done W = F × s = 20 N × 10 m = 200 J.(b) From the work-energy theorem, work done = change in kinetic energy.200 J =

\frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}

Assuming the object starts from rest, u = 0.200 J =

\frac{1}{2} \times 5 kg \times v^{2}

v^{2} = \frac{2 \times 200}{5} = 80

$v = \sqrt{80} \approx 8.94$ m/s.
Acceleration $a = \frac{v^{2} - u^{2}}{2 s} = \frac{80 - 0}{2 \times 10} = \frac{80}{20} = 4 {m/s}^{2}$ .

(a) Both assertion and reason are true, and the reason is the correct explanation of the assertion.

(a) Initial kinetic energy,

K_{i} = \frac{1}{2} m u^{2} = \frac{1}{2} \times 10 kg \times (5 m/s)^{2} = \frac{1}{2} \times 10 \times 25 = 125 J

.(b) Final kinetic energy,

K_{f} = \frac{1}{2} m v^{2} = \frac{1}{2} \times 10 kg \times (10 m/s)^{2} = \frac{1}{2} \times 10 \times 100 = 500 J

.(c) Work done = Change in kinetic energy =

K_{f} - K_{i} = 500 J - 125 J = 375 J

(a) Velocity

v = 54 km/h = 54 \times \frac{1000}{3600} m/s = 15 m/s

.Initial kinetic energy,

K_{i} = \frac{1}{2} m v^{2} = \frac{1}{2} \times 1200 kg \times (15 m/s)^{2} = 600 \times 225 = 135000 J

.(b) Work done by the brakes = Change in kinetic energy =

K_{f} - K_{i} = 0 J - 135000 J = - 135000 J

. (Negative work is done).(c) Work done = Force × Displacement.

- 135000 J = F \times (- 5 m)

(Assuming displacement is in the direction of motion, and force is opposite, hence displacement can be taken as -5m for calculation of work done by force. Or, more simply,

- 135000 J = F \times 5 m

, where F is the magnitude of the braking force).

F = \frac{135000}{5} = 27000 N

(a) Potential energy at the top,

U = m g h = 0.5 kg \times 10 {m/s}^{2} \times 30 m = 150 J

.(b) At the top, the ball is momentarily at rest before rolling down, so its kinetic energy is 0 J.(c) By the principle of conservation of mechanical energy, Potential energy at the top = Kinetic energy at the bottom.

150 J = \frac{1}{2} m v^{2}

150 J = \frac{1}{2} \times 0.5 kg \times v^{2}

v^{2} = \frac{150 \times 2}{0.5} = \frac{300}{0.5} = 600

$v = \sqrt{600} \approx 24.5 m/s$ .
(d) The formula for potential energy ( $m g h$ ) and kinetic energy ( $\frac{1}{2} m v^{2}$ ) both depend on mass, but when equating them for conservation of energy ( $m g h = \frac{1}{2} m v^{2}$ ), the mass ( $m$ ) cancels out from both sides, leaving $g h = \frac{1}{2} v^{2}$ , which means the velocity depends only on height and 'g'.

(a) Work done against gravity = Force × Displacement =

m g \times h = 5 kg \times 10 {m/s}^{2} \times 10 m = 500 J

.(b) The potential energy gained is equal to the work done against gravity, which is 500 J.

Initial kinetic energy

K_{A} = \frac{1}{2} m_{A} v_{A}^{2}

and

K_{B} = \frac{1}{2} m_{B} v_{B}^{2}

.Given

K_{A} = K_{B}

\frac{1}{2} m_{A} v_{A}^{2} = \frac{1}{2} m_{B} v_{B}^{2}

m_{A} v_{A}^{2} = m_{B} v_{B}^{2}

Substitute $m_{A} = 4 m_{B}$ :

4 m_{B} v_{A}^{2} = m_{B} v_{B}^{2}

4 v_{A}^{2} = v_{B}^{2}

2 v_{A} = v_{B}

The ratio of their velocities, $v_{A} / v_{B} = v_{A} / (2 v_{A}) = 1 / 2$ .

(a) Both assertion and reason are true, and the reason is the correct explanation of the assertion.

(a) Initial kinetic energy,

K_{i} = \frac{1}{2} m v^{2} = \frac{1}{2} \times 0.5 kg \times (2 m/s)^{2} = \frac{1}{2} \times 0.5 \times 4 = 1 J

.(b) Potential energy gained,

U = m g h = 0.5 kg \times 10 {m/s}^{2} \times 0.2 m = 1 J

.(c) Energy lost due to friction = Initial kinetic energy - Potential energy gained = 1 J - 1 J = 0 J. (This implies no energy loss due to friction in this specific scenario as described by the values).*Correction based on typical scenario: If the car comes to rest, its final KE is 0. So, Initial KE = Work done by friction + Potential energy gained.

1 J = Work done by friction + 1 J

. This would mean work done by friction is 0 J. This implies no energy loss. If the problem intended energy loss, the height achieved would be less than the initial energy. Let's re-evaluate with a realistic scenario for 'rest'.If the car comes to rest at height h, then its final KE is 0.Initial Energy (KE) = Work done by friction + Final Potential Energy

1 J = Work done by friction + U

If the question implies that 0.2m is the maximum height it reaches, then the $U$ at that height is 1 J. This means the initial KE was 1 J. If the car comes to rest, all its initial kinetic energy must be converted into potential energy and work done against friction.

K_{i n i t i a l} = U_{f i n a l} + Work done by friction

1 J = 1 J + Work done by friction

Work done by friction = 0 J. This suggests ideal conditions.

Mass of water = Volume × Density = 5000 litres × 1000 kg/m³ = 5 m³ × 1000 kg/m³ = 5000 kg.Work done =

m g h = 5000 kg \times 10 {m/s}^{2} \times 20 m = 1000000 J

.Time taken = 1 minute = 60 seconds.Power =

\frac{Work done}{Time taken} = \frac{1000000 J}{60 s} \approx 16666.67 W

Work done =

m g h = 200 kg \times 10 {m/s}^{2} \times 1.5 m = 3000 J

.Time taken = 5 seconds.Power =

\frac{Work done}{Time taken} = \frac{3000 J}{5 s} = 600 W

(a) Both assertion and reason are true, and the reason is the correct explanation of the assertion.

(a) Work done against gravity (Load) =

m g h = 10 kg \times 10 {m/s}^{2} \times 5 m = 500 J

.(b) Work done by effort = Effort × Distance moved by effort.Mechanical Advantage (MA) = Load / Effort = 500 J / 50 N = 10. So, the distance moved by effort should be 10 times the height to have MA=10.If MA = Load / Effort, and assuming Load is the weight (500 N), MA = 500 N / 50 N = 10.Work done by effort = Effort × Distance moved by effort. Let the distance be

d_{e}

.Work done by effort = 50 N ×

d_{e}

.If we assume ideal conditions where work done by effort = work done on load:

50 N \times d_{e} = 500 J

$d_{e} = \frac{500}{50} = 10 m$ .
Work done by the effort = 50 N × 10 m = 500 J. (Note: If friction is present, work by effort will be > 500 J).
(c) Mechanical Advantage (MA) = Load / Effort = 500 N / 50 N = 10.
(d) A movable pulley system or a block and tackle system can have a mechanical advantage greater than 1. A fixed pulley has MA=1. Since MA=10, it is likely a movable pulley or a block and tackle system where the number of supporting ropes for the load is 10.

(a) At the highest point, the child momentarily stops before swinging back, so the energy is predominantly potential energy.(b) At the lowest point, the child is moving with maximum speed, so the energy is predominantly kinetic energy.(c) The principle of conservation of mechanical energy describes the transformation of energy between the highest and lowest points, where potential energy is converted into kinetic energy and vice versa, assuming no energy losses.(d) In reality, friction at the pivot point of the swing and air resistance cause energy loss, which reduces the mechanical energy of the swing. This leads to the swing slowing down and eventually stopping.

(a) Work done by the family = Force × Distance = 500 N × 100 m = 50000 J.(b) Potential energy gained by the car =

m g h = 1000 kg \times 10 {m/s}^{2} \times 10 m = 100000 J

.(c) Assuming the car started from rest, the initial kinetic energy is 0 J.The total energy at the top = Kinetic energy + Potential energy.If we assume the family's work overcomes both the potential energy gain and any friction (which is not given), then:Work done =

Δ K E + Δ P E

50000 J = K E_{f i n a l} + 100000 J

. This gives a negative KE, which is impossible. This indicates an error in the problem statement or interpretation. Let's assume the 500N is the net force along the incline after considering friction, or the problem implicitly means to find the energy gained and kinetic energy based on the forces.Re-interpreting: If the car starts from rest, and moves up a distance of 100m along the incline gaining a height of 10m, and the family applies a force of 500N.Work done by family = 500 N * 100 m = 50000 J.Potential Energy gained =

m g h = 1000 \times 10 \times 10 = 100000 J

.If the car starts from rest, and reaches the top with some KE, then Work done by family = Change in KE + Change in PE.

50000 = K E_{f i n a l} - 0 + 100000 - 0

K E_{f i n a l} = 50000 - 100000 = - 50000 J

. This is impossible.

Let's assume the 500 N is the net force that causes motion up the incline, and there's some force opposing it (like friction or gravity component).
Component of gravity along the incline = $m g \sin θ$ . Since $\sin θ = \frac{h}{L} = \frac{10}{100} = 0.1$ , this force is $1000 \times 10 \times 0.1 = 1000 N$ .
If the family applies 500N, and gravity component is 1000N, they cannot move it up.
Revised problem interpretation for a solvable scenario: Assume the family applies a force in addition to gravity, or the problem intended a different force value. Or, perhaps the question implies that 500N is the net force exerted by the family and other forces like gravity component and friction oppose this.

Let's assume the question meant: A force of 1500 N is applied along the incline.
Work done by family = 1500 N × 100 m = 150000 J.
Potential Energy gained = 100000 J.
Work done = $Δ K E + Δ P E$

150000 = K E_{f i n a l} - 0 + 100000 - 0

$K E_{f i n a l} = 150000 - 100000 = 50000 J$ .

If we strictly use the given values:
Work done by the family = 500 N × 100 m = 50000 J.
Potential energy gained = 100000 J.
Since Work done by the family (50000 J) is less than the potential energy gained (100000 J), and the car starts from rest, it's impossible for the car to reach the top with this force alone, unless some other force is pushing it or the initial force is much larger. This problem statement seems flawed with the given values.

Assuming the question implies 'what is the KE if the forces allowed it to reach':
If we assume the work done by the family is solely responsible for the increase in KE and PE.
Work done = $W_{f a m i l y} = 500 N \times 100 m = 50000 J$ .
Potential Energy gained = $m g h = 1000 kg \times 10 {m/s}^{2} \times 10 m = 100000 J$ .
By work-energy theorem: $W_{n e t} = Δ K E$ .
If we consider the force applied by the family and the component of gravity along the incline and friction as the forces doing work.
Let's assume the 500N is the net force available to increase KE.
Then, $Δ K E = 500 N \times 100 m = 50000 J$ .
If it starts from rest, the final KE is 50000 J.

Given the typical structure of such problems, and the likely intent:
(a) Work done by the family = 500 N × 100 m = 50000 J.
(b) Potential energy gained = 100000 J.
(c) This problem cannot be solved as stated because the work done by the family (50000 J) is less than the potential energy gained (100000 J) if starting from rest and assuming ideal conditions. If we assume the 500N is a net force that leads to motion, then the kinetic energy gained would be 50000 J.

(a) Work done by the applied force = Force × Distance = 50 N × 10 m = 500 J.(b) Work done by friction = Force × Distance = 20 N × 10 m = 200 J. Since friction acts opposite to motion, this work is negative. So, Work done by friction = -200 J.(c) Net work done = Work done by applied force + Work done by friction = 500 J + (-200 J) = 300 J.(d) Change in kinetic energy = Net work done = 300 J.

(a) The potential energy of the water stored at a high level is converted into kinetic energy as it flows down. This kinetic energy of the water is then used to rotate the turbines, which in turn drive generators to produce electrical energy. Thus, the transformation is: Potential Energy

\to

Kinetic Energy

\to

Mechanical Energy of Turbine

\to

Electrical Energy.(b) Potential energy =

m g h = 2000 kg \times 10 {m/s}^{2} \times 50 m = 1000000 J

.(c) Assuming all potential energy is converted to kinetic energy,

P E = K E

1000000 J = \frac{1}{2} m v^{2}

1000000 J = \frac{1}{2} \times 2000 kg \times v^{2}

1000000 = 1000 v^{2}

v^{2} = 1000

$v = \sqrt{1000} \approx 31.6 m/s$ .

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Work, Energy, and Simple Machines — Competency Based Questions

Competency based questions

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