NCERT Solutions for Class 9 Science Exploration Chapter 7: Work, Energy and Simple Machines
Session 2026-27 Updated
Revise, Reflect, Refine
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.
(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. Explanation: Work is done only when force causes displacement. If there is no movement, no work is done.
(ii) True. Explanation: The force applied and displacement are in the same direction, so work done is positive.
(iii) True. Explanation: Both work and energy are measured in joules (J) in SI units.
(iv) False. Explanation: A motionless object has no kinetic energy; a stretched rubber band has potential energy.
(v) True. Explanation: Energy can be converted from one form to another (law of conservation of energy).
(i) False. Explanation: Work is done only when force causes displacement. If there is no movement, no work is done.
(ii) True. Explanation: The force applied and displacement are in the same direction, so work done is positive.
(iii) True. Explanation: Both work and energy are measured in joules (J) in SI units.
(iv) False. Explanation: A motionless object has no kinetic energy; a stretched rubber band has potential energy.
(v) True. Explanation: Energy can be converted from one form to another (law of conservation of energy).
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.
(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.
Answer:
(i) Work done = Force × displacement (in the direction of force).
(ii) 1 joule of work is done when a force of 1 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 ½mv².
(iv) The potential energy of an object of mass m at a small height h from the Earth’s surface is mgh.
(v) Power is defined as the rate at which work is done.
(i) Work done = Force × displacement (in the direction of force).
(ii) 1 joule of work is done when a force of 1 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 ½mv².
(iv) The potential energy of an object of mass m at a small height h from the Earth’s surface is mgh.
(v) Power is defined as the rate at which work is done.
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.
(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: Correct statements: (iii) and (iv).
Explanation: Force is not zero (gravity acts downward), and acceleration is not zero (g acts downward). However, velocity becomes zero ⇒ KE = 0, and height is maximum ⇒ PE is 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 firecracker
(vii) Speaking into a microphone
(viii) A glowing electric bulb
(ix) A solar panel
(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
(ix) A solar panel
Answer:
(i) Kinetic energy ⇒ Potential energy: When a truck moves uphill, it gains height, converting kinetic energy into gravitational potential energy.
(ii) Potential energy ⇒ Kinetic energy: A wound spring stores potential energy, which is used to produce motion when it unwinds.
(iii) Light energy ⇒ Chemical energy: Plants use sunlight to prepare food, converting light energy into stored chemical energy.
(iv) Potential energy ⇒ Kinetic energy: Water stored at a height has potential energy that changes into motion as it flows downward.
(v) Chemical energy ⇒ Heat energy + Light energy: The chemical energy stored in the matchstick is released as heat and light when burned.
(vi) Chemical energy ⇒ Heat + Light + Sound + Kinetic energy: Firecrackers contain chemical energy suddenly released in multiple forms.
(vii) Sound energy ⇒ Electrical energy: Sound energy is produced and converted into electrical signals by the microphone.
(viii) Electrical energy ⇒ Light energy + Heat energy: Electrical energy supplied is converted into light and heat.
(ix) Light energy ⇒ Electrical energy: Solar panels convert sunlight directly into electrical energy.
(i) Kinetic energy ⇒ Potential energy: When a truck moves uphill, it gains height, converting kinetic energy into gravitational potential energy.
(ii) Potential energy ⇒ Kinetic energy: A wound spring stores potential energy, which is used to produce motion when it unwinds.
(iii) Light energy ⇒ Chemical energy: Plants use sunlight to prepare food, converting light energy into stored chemical energy.
(iv) Potential energy ⇒ Kinetic energy: Water stored at a height has potential energy that changes into motion as it flows downward.
(v) Chemical energy ⇒ Heat energy + Light energy: The chemical energy stored in the matchstick is released as heat and light when burned.
(vi) Chemical energy ⇒ Heat + Light + Sound + Kinetic energy: Firecrackers contain chemical energy suddenly released in multiple forms.
(vii) Sound energy ⇒ Electrical energy: Sound energy is produced and converted into electrical signals by the microphone.
(viii) Electrical energy ⇒ Light energy + Heat energy: Electrical energy supplied is converted into light and heat.
(ix) Light energy ⇒ Electrical energy: Solar panels convert sunlight directly into electrical energy.
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 s♠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?
(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) Gain in gravitational potential energy is: PE = mgh. Substituting values: PE = 50 × 10 × 72.5 = 36250 J.
(ii) Gain in potential energy will be the same as in part (i), because the height reached is the same: 36,250 J. Potential energy depends only on mass, gravity, and height, not on how the height is reached.
(iii) Conclusion: The potential energy does not depend on the path taken. It depends only on the initial and final positions (height). Both the elevator and stairs result in the same potential energy gain.
(i) Gain in gravitational potential energy is: PE = mgh. Substituting values: PE = 50 × 10 × 72.5 = 36250 J.
(ii) Gain in potential energy will be the same as in part (i), because the height reached is the same: 36,250 J. Potential energy depends only on mass, gravity, and height, not on how the height is reached.
(iii) Conclusion: The potential energy does not depend on the path taken. It depends only on the initial and final positions (height). Both the elevator and stairs result in the same potential energy gain.
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 of each floor = h.
Height to 10th floor = 10h, Height to 20th floor = 20h.
Using Energy = mgh:
For 10th floor: E₁ = mg(10h) = 10mgh.
For 20th floor: E₂ = mg(20h) = 20mgh.
So, E₂ = 2E₁. The energy required is double.
Using Power = Work / Time:
Let time for 10th floor = t, Time for 20th floor = 2t.
P₁ = E₁ / t = 10mgh / t.
P₂ = E₂ / 2t = 20mgh / 2t = 10mgh / t.
So, P₂ = P₁. The power required remains the same.
Let height of each floor = h.
Height to 10th floor = 10h, Height to 20th floor = 20h.
Using Energy = mgh:
For 10th floor: E₁ = mg(10h) = 10mgh.
For 20th floor: E₂ = mg(20h) = 20mgh.
So, E₂ = 2E₁. The energy required is double.
Using Power = Work / Time:
Let time for 10th floor = t, Time for 20th floor = 2t.
P₁ = E₁ / t = 10mgh / t.
P₂ = E₂ / 2t = 20mgh / 2t = 10mgh / t.
So, P₂ = P₁. The power required remains the same.
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:
Factors determining energy: The energy required equals the gain in gravitational potential energy, PE = mgh. The factors are mass of the flag (m), height of the flagpole (h), and acceleration due to gravity (g).
Effect of speed on work done: Raising the flag slowly or quickly does NOT change the amount of work done. Work depends only on force and displacement, not on time or speed.
Effect on power: Power is defined as Power = Work / Time. If speed is doubled, the time taken becomes half. New power = Work / (Time/2) = 2 × (Work/Time). Therefore, the power required doubles.
Factors determining energy: The energy required equals the gain in gravitational potential energy, PE = mgh. The factors are mass of the flag (m), height of the flagpole (h), and acceleration due to gravity (g).
Effect of speed on work done: Raising the flag slowly or quickly does NOT change the amount of work done. Work depends only on force and displacement, not on time or speed.
Effect on power: Power is defined as Power = Work / Time. If speed is doubled, the time taken becomes half. New power = Work / (Time/2) = 2 × (Work/Time). Therefore, the power required doubles.
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:
Fuel used ∝ energy required ∝ Kinetic Energy (KE = ½mv²). Since final velocity is the same, KE depends only on total mass.
Day 1: Total mass = 60 kg + 100 kg = 160 kg. KE₁ = ½ × 160 × v².
Day 2: Total mass = 60 kg + 40 kg + 100 kg = 200 kg. KE₂ = ½ × 200 × v².
Ratio of fuel used: KE₁ : KE₂ = 160 : 200 = 4 : 5. The ratio of fuel used on the two days is 4 : 5.
Fuel used ∝ energy required ∝ Kinetic Energy (KE = ½mv²). Since final velocity is the same, KE depends only on total mass.
Day 1: Total mass = 60 kg + 100 kg = 160 kg. KE₁ = ½ × 160 × v².
Day 2: Total mass = 60 kg + 40 kg + 100 kg = 200 kg. KE₂ = ½ × 200 × v².
Ratio of fuel used: KE₁ : KE₂ = 160 : 200 = 4 : 5. The ratio of fuel used on the two days is 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.
Answer:
For a balanced seesaw, clockwise moment = anticlockwise moment.
Using Moment = Force × Distance from fulcrum.
Let Weight of child = W, Weight of adult = 2W. Let Distance of child = d₁, Distance of adult = d₂.
For balance: W × d₁ = 2W × d₂ ⇒ d₁ = 2d₂.
Conclusion: The child must sit at twice the distance from the fulcrum compared to the adult. For the seesaw to remain balanced, the lighter child sits farther and the heavier adult sits closer.
For a balanced seesaw, clockwise moment = anticlockwise moment.
Using Moment = Force × Distance from fulcrum.
Let Weight of child = W, Weight of adult = 2W. Let Distance of child = d₁, Distance of adult = d₂.
For balance: W × d₁ = 2W × d₂ ⇒ d₁ = 2d₂.
Conclusion: The child must sit at twice the distance from the fulcrum compared to the adult. For the seesaw to remain balanced, the lighter child sits farther and the heavier adult sits closer.
10. A ball of mass 2 kg is thrown up with a velocity of 20 m s⁻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 m s♠²)?
(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 s♠²)?
Answer:
(i) Upward motion: Work done by gravity is negative (force opposite to motion). Downward motion: Work done by gravity is positive (force in direction of motion).
(ii) Initial KE = ½mv² = ½ × 2 × (20)² = 400 J.
Potential energy at height: PE = mgh = 2 × 10 × 19.4 = 388 J.
Work done by air resistance = Change in mechanical energy = Final energy − Initial energy = 388 J − 400 J = −12 J.
(i) Upward motion: Work done by gravity is negative (force opposite to motion). Downward motion: Work done by gravity is positive (force in direction of motion).
(ii) Initial KE = ½mv² = ½ × 2 × (20)² = 400 J.
Potential energy at height: PE = mgh = 2 × 10 × 19.4 = 388 J.
Work done by air resistance = Change in mechanical energy = Final energy − Initial energy = 388 J − 400 J = −12 J.
11. A 10.0 kg block is moving on a horizontal floor with negligible friction. 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?
Answer:
Given: Mass = 10 kg, Initial KE = 180 J.
(i) Speed at 0 m: Using KE = ½mv² ⇒ 180 = ½ × 10 × v² ⇒ v² = 36 ⇒ v = 6 m/s.
(ii) Speed at 4 m: Work done = Area under force-displacement graph.
0–1 m area = ½ × 1 × 50 = 25 J.
1–3 m area = 2 × 50 = 100 J.
3–4 m area = ½ × 1 × 50 = 25 J.
Total work = 25 + 100 + 25 = 150 J.
Final KE = Initial KE + Work = 180 + 150 = 330 J.
Using 330 = ½ × 10 × v² ⇒ v² = 66 ⇒ v = √66 ≈ 8.12 m/s.
Negative acceleration: No, because the force is always in the direction of motion.
Given: Mass = 10 kg, Initial KE = 180 J.
(i) Speed at 0 m: Using KE = ½mv² ⇒ 180 = ½ × 10 × v² ⇒ v² = 36 ⇒ v = 6 m/s.
(ii) Speed at 4 m: Work done = Area under force-displacement graph.
0–1 m area = ½ × 1 × 50 = 25 J.
1–3 m area = 2 × 50 = 100 J.
3–4 m area = ½ × 1 × 50 = 25 J.
Total work = 25 + 100 + 25 = 150 J.
Final KE = Initial KE + Work = 180 + 150 = 330 J.
Using 330 = ½ × 10 × v² ⇒ v² = 66 ⇒ v = √66 ≈ 8.12 m/s.
Negative acceleration: No, because the force is always in the direction of motion.
12. The gravitational attraction on the surface of the Moon is about 1/6th 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:
Using the relation: h = u² / (2g). For the same initial velocity (u), height is inversely proportional to gravity: h ∝ 1/g.
Given height on Earth = 8 m, and gravity on Moon is 1/6th of Earth’s gravity.
Since gravity on the Moon is 1/6th, the height reached will be 6 times greater.
Height on Moon: hᵒ = 6 × 8 = 48 m.
The ball will rise up to a height of 48 m on the Moon.
Using the relation: h = u² / (2g). For the same initial velocity (u), height is inversely proportional to gravity: h ∝ 1/g.
Given height on Earth = 8 m, and gravity on Moon is 1/6th of Earth’s gravity.
Since gravity on the Moon is 1/6th, the height reached will be 6 times greater.
Height on Moon: hᵒ = 6 × 8 = 48 m.
The ball will rise up to a height of 48 m on the Moon.
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.
(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?
(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 with a constant speed of 35 m s⁻¹. This means the car is in uniform motion and no acceleration acts on it during this part of the motion.
(ii) Given m = 1000 kg and v = 35 m/s. KE = ½mv² = ½ × 1000 × (35)² = 500 × 1225 = 612500 J.
(iii) Work done by brakes equals the change in kinetic energy. Initial KE at B = 6,12,500 J, Final KE at C = 0 J.
Work done = 0 − 612500 = −612500 J. The negative sign shows the braking force acts opposite to motion.
(iv) The kinetic energy transforms mainly into heat energy due to friction between brake pads/wheels and tyres/road, and a small part into sound energy.
(i) Between A and B, the car moves with a constant speed of 35 m s⁻¹. This means the car is in uniform motion and no acceleration acts on it during this part of the motion.
(ii) Given m = 1000 kg and v = 35 m/s. KE = ½mv² = ½ × 1000 × (35)² = 500 × 1225 = 612500 J.
(iii) Work done by brakes equals the change in kinetic energy. Initial KE at B = 6,12,500 J, Final KE at C = 0 J.
Work done = 0 − 612500 = −612500 J. The negative sign shows the braking force acts opposite to motion.
(iv) The kinetic energy transforms mainly into heat energy due to friction between brake pads/wheels and tyres/road, and a small part into sound energy.
14. The potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown. At O, the velocity of the ball is 0 m s⁻¹ and potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.
Answer:
At O: Velocity = 0 m s⁻¹, PE = 30 J. Total mechanical energy = PE + KE = 30 + 0 = 30 J. Since the track is frictionless, total mechanical energy remains constant at 30 J.
At P: PE = 20 J. KE = 30 − 20 = 10 J. Using KE = ½mv² ⇒ 10 = ½ × 0.5 × v² ⇒ v² = 40 ⇒ v = √40 ≈ 6.32 m/s.
At Q: PE = 30 J. KE = 30 − 30 = 0 J. Thus, v = 0 m/s.
At R: PE = 40 J. This is greater than the total mechanical energy (30 J). Hence, the ball cannot reach point R. Velocity at R cannot be calculated.
At O: Velocity = 0 m s⁻¹, PE = 30 J. Total mechanical energy = PE + KE = 30 + 0 = 30 J. Since the track is frictionless, total mechanical energy remains constant at 30 J.
At P: PE = 20 J. KE = 30 − 20 = 10 J. Using KE = ½mv² ⇒ 10 = ½ × 0.5 × v² ⇒ v² = 40 ⇒ v = √40 ≈ 6.32 m/s.
At Q: PE = 30 J. KE = 30 − 30 = 0 J. Thus, v = 0 m/s.
At R: PE = 40 J. This is greater than the total mechanical energy (30 J). Hence, the ball cannot reach point R. Velocity at R cannot be calculated.
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 s♠².
(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 s♠².
Answer:
(i) Given: m = 1.5 kg, h = 10 m, u = 0, g = 10 m/s².
Using v² = u² + 2gh ⇒ v² = 0 + 2 × 10 × 10 = 200 ⇒ v = √200 ≈ 14.14 m/s.
(ii) Just before striking the sand, KE = loss in PE = mgh = 1.5 × 10 × 10 = 150 J.
This entire energy is used against the resistive force of sand: Work done = Force × distance.
150 = 3000 × d ⇒ d = 150 / 3000 = 0.05 m.
Converting into centimetres: 0.05 m = 5 cm depth.
(i) Given: m = 1.5 kg, h = 10 m, u = 0, g = 10 m/s².
Using v² = u² + 2gh ⇒ v² = 0 + 2 × 10 × 10 = 200 ⇒ v = √200 ≈ 14.14 m/s.
(ii) Just before striking the sand, KE = loss in PE = mgh = 1.5 × 10 × 10 = 150 J.
This entire energy is used against the resistive force of sand: Work done = Force × distance.
150 = 3000 × d ⇒ d = 150 / 3000 = 0.05 m.
Converting into centimetres: 0.05 m = 5 cm depth.
Very Short Answer Type Questions
1. Define work done by a constant force.
Answer: Work done by a constant force on an object = force applied × displacement in the direction of force. Formula: W = F × s. SI unit is joule (J), and work has no direction.
2. When is work done on an object equal to zero?
Answer: Work done is zero when: (i) force is zero, (ii) displacement is zero (e.g., pushing a wall) or (iii) force acts perpendicular to displacement (e.g., carrying a box horizontally while walking).
3. Define 1 joule of work.
Answer: One joule of work is done when a constant force of 1 newton displaces an object by 1 metre in the direction of the force. Mathematically: 1 J = 1 N × 1 m = 1 kg m² s².
4. What is kinetic energy? Give its formula.
Answer: Kinetic energy is the energy possessed by an object due to its motion. Formula: K = ½mv², where m is mass in kg and v is velocity in m/s. SI unit is joule (J).
5. What is gravitational potential energy? Give its formula.
Answer: Gravitational potential energy is the energy stored in an Earth-object system due to the object’s height above the ground. Formula: U = mgh, where m = mass, g = 9.8 m/s², h = height.
6. State the work-energy theorem.
Answer: Work done on an object equals the change in its energy: W = ΔE. When positive work is done, energy increases; when negative work is done, energy decreases. It holds for variable forces too.
7. Define power and give its SI unit.
Answer: Power is defined as the rate at which work is done: P = W/t. SI unit of power is watt (W), where 1 watt = 1 joule per second (1 J/s). Named after James Watt.
8. What is mechanical energy?
Answer: Mechanical energy is the sum of kinetic energy and potential energy of an object: ME = K + U = ½mv² + mgh. For a freely falling object, mechanical energy remains constant if no friction acts.
9. What is a simple machine?
Answer: A simple machine is a device that makes work easier by changing the magnitude or direction of an applied force. Simple machines do not reduce total work done. Examples: pulley, inclined plane, lever.
10. Define mechanical advantage.
Answer: Mechanical advantage is the ratio of load to effort applied on a machine: MA = load/effort. A mechanical advantage greater than 1 means the machine multiplies the applied force to overcome a larger load.
11. What is the mechanical advantage of a fixed pulley?
Answer: A fixed pulley only changes the direction of the applied force — it does not reduce the magnitude of force needed. Since effort equals load, the mechanical advantage of a fixed pulley is exactly 1.
12. Write the formula for mechanical advantage of an inclined plane.
Answer: Mechanical advantage of an inclined plane = L/h, where L is the length of the inclined surface and h is the vertical height gained. A longer, shallower incline gives a higher mechanical advantage.
13. What is the principle of a lever?
Answer: The lever principle states: effort × effort arm = load × load arm (Or F1 × d1 = F2 × d2). By increasing the effort arm, a smaller effort can overcome a larger load. Mechanical advantage = effort arm/load arm.
14. If velocity of an object doubles, how does its kinetic energy change?
Answer: Kinetic energy K = ½mv². If velocity doubles (2v), KE = ½m(2v)² = 4 × ½mv². Kinetic energy becomes 4 times the original value. KE is proportional to the square of velocity.
15. Name the three classes of levers with one example each.
Answer: Class I: Fulcrum between load and effort — seesaw, scissors. Class II: Load between fulcrum and effort — wheelbarrow, bottle opener. Class III: Effort between fulcrum and load — tweezers, broom.
Short Answer Type Questions
1. Explain why a person pushing a rigid wall does no work on the wall, yet feels tired.
Answer: Scientific work requires both force and displacement. Although the person applies force on the wall, the wall does not move — displacement is zero. Therefore, W = F × s = F × 0 = 0. However, the person’s muscles repeatedly expand and contract internally, consuming the body’s chemical energy. This internal energy expenditure causes tiredness even though no scientific work is done on the wall.
2. Why does a goalkeeper do negative work on the football while stopping it?
Answer: The goalkeeper applies a force on the ball in the direction opposite to the ball’s motion (backward force on a forward-moving ball). Since force and displacement are in opposite directions, the work done by the goalkeeper is negative: W = F × (−s) = negative. This negative work removes energy from the ball, reducing its velocity to zero.
3. Distinguish between kinetic energy and potential energy with one example each.
Answer: Kinetic Energy: Energy possessed by an object due to its motion. Formula: K = ½mv². Example: a moving cricket ball possesses kinetic energy and can knock down wickets. Potential Energy: Energy stored due to an object’s position or deformation. Formula: U = mgh (gravitational). Example: a flowerpot raised to a height stores gravitational potential energy; when it falls, this converts to kinetic energy. Both are measured in joules (J).
4. A child of mass m slides down a frictionless slide of height h. What is the velocity at the bottom? Does it depend on the mass or shape of the slide?
Answer: At the top: PE = mgh, KE = 0. At the bottom: PE = 0, KE = ½mv². By conservation of mechanical energy: mgh = ½mv². Cancelling m from both sides: v = √(2gh). Since mass cancels, the velocity depends only on height h, not on the mass of the child. The shape of the slide also does not matter as long as height h and the absence of friction are the same.
5. Explain conservation of mechanical energy using a simple pendulum.
Answer: At the extreme position P, the pendulum bob has maximum potential energy (mgh) and zero kinetic energy. As it swings to the lowest position Q, PE converts entirely to kinetic energy — KE is maximum and PE is zero. At the other extreme R, KE converts back to PE. Throughout, total mechanical energy (KE + PE) remains constant at mgh, demonstrating conservation of mechanical energy. In reality, friction at the support and air resistance gradually reduce the total mechanical energy, causing the pendulum to eventually stop.
6. How does an inclined plane act as a simple machine? Why does a longer, shallower incline require less effort?
Answer: An inclined plane reduces the force needed to raise a heavy object to a height by spreading the work over a longer distance. Using the work-energy theorem: F’ × L = mgh, so F’ = mgh/L. Mechanical advantage = L/h. A longer inclined plane (larger L) for the same height h means a larger MA, so the effort F’ required is proportionally smaller. The total work done (mgh) remains the same — the machine redistributes force and distance but cannot reduce the work itself.
7. State the three classes of levers with their characteristic feature and two real-life examples of each.
Answer:
Class I — Fulcrum lies between load and effort. The effort and load are on opposite sides of the fulcrum. MA can be greater or less than 1. Examples: seesaw, scissors, crowbar, pliers, balance scale.
Class II — Load lies between fulcrum and effort. Effort arm is always longer than load arm. MA is always greater than 1. Examples: wheelbarrow, bottle opener, lemon squeezer.
Class III — Effort lies between fulcrum and load. Load arm is always longer than effort arm. MA is always less than 1, but speed and range of movement increase. Examples: tweezers, broom, hammer, oar.
Class I — Fulcrum lies between load and effort. The effort and load are on opposite sides of the fulcrum. MA can be greater or less than 1. Examples: seesaw, scissors, crowbar, pliers, balance scale.
Class II — Load lies between fulcrum and effort. Effort arm is always longer than load arm. MA is always greater than 1. Examples: wheelbarrow, bottle opener, lemon squeezer.
Class III — Effort lies between fulcrum and load. Load arm is always longer than effort arm. MA is always less than 1, but speed and range of movement increase. Examples: tweezers, broom, hammer, oar.
8. A man runs up a flight of stairs in 10 seconds. His friend walks up the same stairs in 50 seconds. Who does more work and who has more power?
Answer: Both travel the same height h and have the same mass m (assume). Therefore, work done = mgh is the same for both. Neither does more work than the other. However, Power = W/t. The man takes 10 seconds while his friend takes 50 seconds for the same work. The man’s power = W/10, and his friend’s power = W/50. Therefore, the man expends 5 times more power than his friend, even though total work is equal.
9. Explain energy transformations in a watermill (gharat or panchakki) as described in the chapter.
Answer: In a traditional Himalayan watermill, water stored at a height possesses gravitational potential energy. As water flows downhill through a pipe (A), this potential energy converts to kinetic energy. The fast-moving water strikes a wheel (B), transferring its kinetic energy to set the wheel into rotational motion. This rotational kinetic energy is transmitted via a shaft to drive the grinding stone (C) above, which converts rotational energy to mechanical energy for grinding grain. Modern hydroelectric dams use the same principle.
10. Why do roads on hills wind around in gentle slopes rather than going straight up? Explain using the concept of inclined planes.
Answer: A straight vertical road up a hill would require vehicles to exert a force equal to the full weight of the vehicle (F = mg) to climb — a very large effort. A winding road acts as a long inclined plane with a gentle slope. By increasing the path length L while keeping the height h the same, the mechanical advantage (MA = L/h) increases significantly. This means the engine needs to apply a much smaller force to climb the same height. The total work done (mgh) is the same either way, but the winding road spreads this work over a larger distance, making the effort (force) at any point much smaller.
Long Answer Type Questions
1. Define work done. When is work done zero? Distinguish between positive and negative work with one example each.
Answer: Work done by a constant force on an object is the product of the force and the displacement in the direction of the force: W = F × s. SI unit is joule (J). Work done is zero when: (i) force is zero, (ii) displacement is zero (pushing a rigid wall), or (iii) force is perpendicular to displacement (carrying a box horizontally).
Positive work: Force and displacement are in the same direction. Example — pushing a wheelchair forward; the applied force and displacement both point forward.
Negative work: Force and displacement are in opposite directions. Example — a goalkeeper stopping a football; the force applied on the ball is opposite to its motion.
Positive work: Force and displacement are in the same direction. Example — pushing a wheelchair forward; the applied force and displacement both point forward.
Negative work: Force and displacement are in opposite directions. Example — a goalkeeper stopping a football; the force applied on the ball is opposite to its motion.
2. Derive the expression for kinetic energy using the work-energy theorem. What happens to kinetic energy when velocity doubles?
Answer: Consider an object of mass m starting from rest (u = 0), acted upon by constant force F over displacement s, attaining velocity v.
Using kinematics: v² = u² + 2as → s = v²/2a.
Work done: W = F × s = ma × v²/2a = ½mv².
By the work-energy theorem, this work equals the kinetic energy gained: K = ½mv².
If velocity doubles from v to 2v: New KE = ½m(2v)² = ½m × 4v² = 4 × ½mv².
Kinetic energy becomes 4 times the original. This is because kinetic energy is proportional to the square of velocity (K ∝ v²). This explains why doubling the speed of a vehicle makes it four times harder to stop.
Using kinematics: v² = u² + 2as → s = v²/2a.
Work done: W = F × s = ma × v²/2a = ½mv².
By the work-energy theorem, this work equals the kinetic energy gained: K = ½mv².
If velocity doubles from v to 2v: New KE = ½m(2v)² = ½m × 4v² = 4 × ½mv².
Kinetic energy becomes 4 times the original. This is because kinetic energy is proportional to the square of velocity (K ∝ v²). This explains why doubling the speed of a vehicle makes it four times harder to stop.
3. State and explain the Law of Conservation of Mechanical Energy for a freely falling object. Show that mechanical energy remains constant at every point.
Answer: Law of Conservation of Mechanical Energy: When no external force other than gravity acts on an object, the total mechanical energy (KE + PE) remains constant.
For an object of mass m dropped from height h:
At top: KE = 0, PE = mgh → Total ME = mgh.
After falling for time t (reaching height h’): PE = mgh’ = mgh − ½mg²t².
KE = ½mv² = ½mg²t².
Total ME = mgh − ½mg²t² + ½mg²t² = mgh.
At ground (h = 0): KE = mgh, PE = 0 → Total ME = mgh.
At every point, ME = mgh (constant). The lost potential energy exactly equals the gained kinetic energy, proving conservation of mechanical energy.
For an object of mass m dropped from height h:
At top: KE = 0, PE = mgh → Total ME = mgh.
After falling for time t (reaching height h’): PE = mgh’ = mgh − ½mg²t².
KE = ½mv² = ½mg²t².
Total ME = mgh − ½mg²t² + ½mg²t² = mgh.
At ground (h = 0): KE = mgh, PE = 0 → Total ME = mgh.
At every point, ME = mgh (constant). The lost potential energy exactly equals the gained kinetic energy, proving conservation of mechanical energy.
4. Explain the three simple machines studied in Chapter 7 — pulley, inclined plane, and lever. Give the formula for mechanical advantage of each.
Answer:
Pulley: A wheel with a groove guiding a rope. A fixed pulley changes only the direction of force. Since effort equals load, MA = 1. Movable pulleys have MA > 1 and can lift heavier loads with less effort.
Inclined Plane: A sloped surface used to raise heavy loads to a height. Using work-energy theorem: F’ × L = mgh → MA = L/h. A longer, shallower incline gives greater MA — the force needed decreases but distance increases.
Lever: A rigid bar rotating about a fulcrum. Principle: effort × effort arm = load × load arm. MA = effort arm/load arm. Greater effort arm → smaller effort needed.
Pulley: A wheel with a groove guiding a rope. A fixed pulley changes only the direction of force. Since effort equals load, MA = 1. Movable pulleys have MA > 1 and can lift heavier loads with less effort.
Inclined Plane: A sloped surface used to raise heavy loads to a height. Using work-energy theorem: F’ × L = mgh → MA = L/h. A longer, shallower incline gives greater MA — the force needed decreases but distance increases.
Lever: A rigid bar rotating about a fulcrum. Principle: effort × effort arm = load × load arm. MA = effort arm/load arm. Greater effort arm → smaller effort needed.
5. Define power. Calculate power in two practical examples from the chapter. Why is the unit watt named after James Watt?
Answer: Power is the rate at which work is done: P = W/t. SI unit is watt (W), where 1 W = 1 J/s. The unit honours James Watt, who invented an efficient steam engine capable of generating continuous rotational motion.
Example 1 — Weightlifter: Lifts a 75 kg mass by 2 m in 5 seconds. Work done = mgh = 75 × 10 × 2 = 1500 J. Power = 1500/5 = 300 W.
Example 2 — Car Engine: A 1000 kg car accelerates from rest to 20 m/s in 10 seconds. Work done = ½mv² = ½ × 1000 × 400 = 200000 J. Power = 200000/10 = 20000 W = 20 kW.
Power measures the speed of energy delivery, not the total amount.
Example 1 — Weightlifter: Lifts a 75 kg mass by 2 m in 5 seconds. Work done = mgh = 75 × 10 × 2 = 1500 J. Power = 1500/5 = 300 W.
Example 2 — Car Engine: A 1000 kg car accelerates from rest to 20 m/s in 10 seconds. Work done = ½mv² = ½ × 1000 × 400 = 200000 J. Power = 200000/10 = 20000 W = 20 kW.
Power measures the speed of energy delivery, not the total amount.