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NCERT Class 9 Science Work, Energy, and Simple Machines: Exercise Solutions

Step-by-step working for every question of End-of-chapter exercises in NCERT's Class 9 Science book, chapter “Work, Energy, and Simple Machines”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Exercises Q1

   True or false: (i) work is done when a force is applied even if the object does not move (ii) lifting a bucket straight up does positive work on it (iii) work and energy both have the SI unit joule (iv) a still, stretched rubber band has kinetic energy (v) energy can change from one form to another.

   1. (i) Work is done only when a force causes displacement, so if the object does not move, no work is done.
   2. (ii) Lifting a bucket upward applies force in the same direction as displacement, which is positive work.
   3. (iii) The SI unit for both work and energy is the joule.
   4. (iv) Kinetic energy is due to motion, so a motionless rubber band has zero kinetic energy.
   5. (v) Energy can be transferred and converted from one form to another.

   Answer: (i) false; (ii) true; (iii) true; (iv) false; (v) true

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2. Exercises Q2

   Fill in: (i) work done = ___ × ___ (along the force) (ii) 1 joule of work is done when ___ newton moves an object 1 m along the force (iii) kinetic energy of mass m at velocity v = ___ (iv) potential energy of mass m at small height h = ___ (v) power is the ___ at which work is done.

   1. (i) Work is calculated as the product of force and displacement along the direction of the force.
   2. (ii) One joule is the work done when a force of one newton moves an object by one metre along the force.
   3. (iii) The kinetic energy of an object with mass m moving at velocity v is given by the formula 1/2 mv².
   4. (iv) The potential energy of an object with mass m at a small height h is given by the formula mgh.
   5. (v) Power is defined as the rate at which work is done.

   Answer: (i) force × displacement; (ii) 1; (iii) 1/2 mv²; (iv) mgh; (v) rate

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3. Exercises Q3

   A ball thrown up is at its highest point. Which are correct? (i) the force on it is zero (ii) its acceleration is zero (iii) its kinetic energy is zero (iv) its potential energy is maximum

   1. (i) The force acting on the ball is not zero because gravity continues to pull it downwards.
   2. (ii) The acceleration is not zero as the ball is under the influence of gravitational force.
   3. (iii) The kinetic energy is zero because the ball momentarily stops moving at the highest point.
   4. (iv) The potential energy is maximum because the ball is at its greatest height above the ground.

   Answer: (iii), (iv)

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4. Exercises Q4

   Name the energy change in: (i) a truck moving uphill (ii) a watch spring unwinding (iii) photosynthesis (iv) water flowing from a dam (v) a burning matchstick (vi) a firecracker exploding (vii) speaking into a microphone (viii) a glowing bulb (ix) a solar panel.

   1. (i) Chemical energy in fuel changes into kinetic energy and potential energy as the truck moves up.
   2. (ii) Potential energy stored in the spring changes into kinetic energy as it unwinds.
   3. (iii) Light energy from the sun is converted into chemical energy stored in food.
   4. (iv) Potential energy of the water changes into kinetic energy as it flows down.
   5. (v) Chemical energy in the matchstick changes into heat and light energy during burning.
   6. (vi) Chemical energy in the firecracker changes into heat, light, and sound energy on explosion.
   7. (vii) Sound energy from the voice is converted into electrical energy by the microphone.
   8. (viii) Electrical energy is converted into light and heat energy in the glowing bulb.
   9. (ix) Light energy from the sun is converted into electrical energy by the solar panel.

   Answer: (i) Chemical to kinetic and potential; (ii) Potential to kinetic; (iii) Light to chemical; (iv) Potential to kinetic; (v) Chemical to heat and light; (vi) Chemical to heat, light, and sound; (vii) Sound to electrical; (viii) Electrical to light and heat; (ix) Light to electrical.

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5. Exercises Q5

   A 50 kg student goes to the top of a 72.5 m building, once by lift and once by the stairs (g = 10 m s⁻²). Find the gain in potential energy (i) by lift (ii) by stairs. (iii) What does this say about potential energy and path?

   1. The gain in potential energy is calculated using the formula PE = mgh.
   2. Here, mass m = 50 kg, height h = 72.5 m, and g = 10 m s⁻².
   3. (i) By lift: PE = 50 × 10 × 72.5 = 36250 J.
   4. (ii) By stairs: PE = 50 × 10 × 72.5 = 36250 J.
   5. (iii) The potential energy gained is the same in both cases because it depends only on the vertical height and not on the path taken.

   Answer: (i) 36250 J; (ii) 36250 J; (iii) Potential energy is independent of the path taken.

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6. Exercises Q6

   A crane lifts a mass to the 10th floor in some time, and the same mass to the 20th floor in double the time (equal floors). How much more energy and power are needed?

   1. Work done to lift the mass is equal to the potential energy gained, which is calculated as mass times gravity times height.
   2. Lifting to the 20th floor requires twice the height compared to the 10th floor, so the energy required is doubled.
   3. Power is defined as the work done divided by the time taken to do that work.
   4. Since the time taken to lift to the 20th floor is double the time for the 10th floor, the power remains the same.
   5. Thus, the energy required is twice as much, but the power required is unchanged.

   Answer: Twice the energy is required, but the power required is the same.

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7. Exercises Q7

   What decides the energy needed to raise a flag up a tall pole with a pulley? Does raising it slowly or quickly change the work? If the speed doubles, how does the power change?

   1. The energy required to raise the flag is determined by its mass and the height of the pole, as this work converts into gravitational potential energy.
   2. The amount of work done remains the same whether the flag is raised slowly or quickly because work depends only on the force and the vertical displacement.
   3. If the speed of raising the flag is doubled, the time taken to reach the top is halved, which causes the power requirement to double.

   Answer: Energy depends on mass and height; work is unchanged by speed; doubling the speed doubles the power.

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8. Exercises Q8

   A 60 kg man rides a 100 kg scooter up to speed v; the next day his 40 kg son rides too, and the same speed is reached in the same time. Find the ratio of fuel used on the two days (no losses).

   1. Mass of man is 60 kg and mass of scooter is 100 kg.
   2. Total mass on the first day is 60 kg + 100 kg = 160 kg.
   3. Mass of son is 40 kg, so total mass on the second day is 60 kg + 40 kg + 100 kg = 200 kg.
   4. Kinetic energy is given by the formula KE = 1/2 × mass × velocity².
   5. Since the speed v and time are the same for both days, the kinetic energy is directly proportional to the total mass.
   6. The ratio of fuel used is equal to the ratio of kinetic energies, which is 160 : 200.
   7. Simplifying the ratio 160 : 200 by dividing both by 40 gives 4 : 5.

   Answer: 4:5

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9. Exercises Q9

   On a balanced seesaw with sliding seats, an adult who weighs twice as much as a child sits on one side and the child on the other. Draw the situation showing their distances from the fulcrum.

   1. i) The seesaw acts as a lever where the balance condition is that the product of weight and distance from the fulcrum is equal on both sides.
   2. ii) Let the weight of the child be W and the adult be 2W, with their distances from the fulcrum being d_child and d_adult respectively.
   3. iii) For the seesaw to be balanced, the equation W × d_child must equal 2W × d_adult.
      why?

      A balanced equation has the same number of atoms of each element on both sides, because atoms are neither created nor destroyed in a chemical reaction (law of conservation of mass). Only the numbers in front (coefficients) change, never the formulas.

   4. iv) Solving this relationship shows that d_child must be twice d_adult, meaning the lighter child sits farther away.
   5. v) The drawing should show the adult closer to the fulcrum and the child at a distance twice that of the adult.

   Answer: The adult sits at a distance d from the fulcrum, and the child sits at a distance 2d from the fulcrum on the opposite side.

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10. Exercises Q10

    A 2 kg ball is thrown up at 20 m s⁻¹. (i) What is the sign of gravity's work going up and coming down? (ii) If it rises only 19.4 m, how much work did air resistance do (g = 10 m s⁻²)?

    1. (i) When the ball moves up, gravity acts downward, so the angle between force and displacement is 180°.
    2. (i) Work done by gravity going up is negative.
    3. (i) When the ball moves down, gravity acts downward and displacement is downward, so the angle is 0°.
    4. (i) Work done by gravity coming down is positive.
    5. (ii) Initial kinetic energy is ½ × 2 × 20² = 400 J.
    6. (ii) Potential energy gained at 19.4 m is mgh = 2 × 10 × 19.4 = 388 J.
    7. (ii) The difference in energy is due to work done against air resistance.
    8. (ii) Work done by air resistance is 388 − 400 = −12 J.

    Answer: (i) Negative going up, positive coming down; (ii) −12 J

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11. Exercises Q11

    A 10.0 kg block on a frictionless floor has 180 J of kinetic energy at 0 m. A force along its motion (Fig. 7.37) rises from 0 to 50 N over 0-1 m, stays 50 N from 1 m to 3 m, and falls to 0 at 4 m. Find its speed at 0 m and at 4 m. Does it ever have negative acceleration?

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. Given mass m = 10.0 kg and initial kinetic energy KE₀ = 180 J at x = 0 m.
    2. Using KE = ½mv², we have 180 = ½ × 10 × v₀², so v₀² = 36 and v₀ = 6 m s⁻¹.
    3. The work done by the force from 0 to 4 m equals the area under the F-x graph.
    4. Area from 0 to 1 m is a triangle: ½ × 1 × 50 = 25 J.
    5. Area from 1 to 3 m is a rectangle: 2 × 50 = 100 J.
    6. Area from 3 to 4 m is a triangle: ½ × 1 × 50 = 25 J.
    7. Total work W = 25 + 100 + 25 = 150 J.
    8. Final kinetic energy KE₄ = KE₀ + W = 180 + 150 = 330 J.
    9. Using KE₄ = ½mv₄², we have 330 = ½ × 10 × v₄², so v₄² = 66 and v₄ = √66 m s⁻¹.
    10. The force is always positive (in the direction of motion), so acceleration a = F/m is always positive.
    11. Therefore, the block never has negative acceleration.

    Answer: Speed at 0 m is 6 m s⁻¹; speed at 4 m is √66 m s⁻¹; no, it never has negative acceleration.

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12. Exercises Q12

    Gravity on the Moon is about 1/6 of that on Earth. A ball thrown up reaches 8 m on Earth. How high does it go on the Moon with the same launch speed?

    1. On Earth, the maximum height reached by the ball is h_E = 8 m.
    2. The acceleration due to gravity on the Moon is g_M = (1/6) g_E.
    3. Using the relation h = u² / (2g), the height is inversely proportional to g.
    4. Therefore, h_M = h_E × (g_E / g_M).
    5. Substituting the values, h_M = 8 m × 6 = 48 m.

    Answer: 48 m

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13. Exercises Q13

    A 1000 kg car moves at a steady 35 m s⁻¹ from A to B (0-1 s), then brakes uniformly to rest at C (3 s) (Fig. 7.38). (i) Describe the motion from A to B. (ii) Find its kinetic energy at A. (iii) Find the work done by the brakes from B to C. (iv) What does the kinetic energy turn into?

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. (i) The car moves with uniform velocity of 35 m s⁻¹ from A to B as its speed remains constant.
    2. (ii) The kinetic energy at A is calculated using the formula KE = ½mv².
    3. (ii) Substituting the values, KE = ½ × 1000 × (35)² = 500 × 1225 = 612500 J.
    4. (iii) The work done by the brakes is equal to the change in kinetic energy from B to C.
    5. (iii) At B, the kinetic energy is 612500 J, and at C, the car is at rest so KE is 0 J.
    6. (iii) Work done = Final KE − Initial KE = 0 − 612500 = −612500 J.
    7. (iv) The kinetic energy is converted into heat energy due to friction between the brakes and the wheels.

    Answer: (i) Uniform motion; (ii) 612500 J; (iii) -612500 J; (iv) Heat energy

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14. Exercises Q14

    A 0.5 kg ball on a frictionless track starts at O at rest with 30 J of potential energy. From Fig. 7.39 its potential energy is 20 J at P, 30 J at Q and 40 J at R. Find its speed at P, Q and R.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. Total mechanical energy is conserved as the track is frictionless, so E = 30 J at all points.
    2. At point P, the potential energy is 20 J, so the kinetic energy is 30 J − 20 J = 10 J.
    3. Using the formula ½mv² = 10, we get 0.5 × 0.5 × v² = 10, which simplifies to 0.25v² = 10.
    4. Solving for v gives v² = 40, so the speed at P is √40 = 2√10 m s⁻¹.
    5. At point Q, the potential energy is 30 J, so the kinetic energy is 30 J − 30 J = 0 J.
    6. Since the kinetic energy is zero, the speed at Q is 0 m s⁻¹.
    7. At point R, the potential energy is 40 J, which is greater than the total energy of 30 J.
    8. The ball cannot reach point R because it does not have enough energy to overcome the potential energy barrier.

    Answer: Speed at P is 2√10 m s⁻¹; speed at Q is 0 m s⁻¹; the ball does not reach R.

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15. Exercises Q15

    A 1.5 kg coconut falls 10 m onto wet sand and stops (g = 10 m s⁻²). (i) Find its speed just before impact. (ii) If the sand's average resistive force is 3000 N and all its energy makes the dent, how deep is the dent?

    1. (i) The potential energy at the top is converted into kinetic energy just before impact.
    2. Using the formula v = √(2gh), substitute g = 10 m s⁻² and h = 10 m.
    3. v = √(2 × 10 × 10) = √200 = 10√2 m s⁻¹.
    4. (ii) The work done by the resistive force of the sand equals the kinetic energy of the coconut.
    5. Kinetic energy KE = ½mv² = ½ × 1.5 × (10√2)² = 0.75 × 200 = 150 J.
    6. Work done by sand = Force × depth, so 3000 × d = 150.
    7. Solving for depth d gives d = 150 ÷ 3000 = 0.05 m.

    Answer: (i) 10√2 m s⁻¹; (ii) 0.05 m

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