NCERT Solutions Class 9 Science Ch 7 Energy 2026-27

At the top of the slide the child has gravitational potential energy mgh. As the child slides down, this changes into kinetic energy. With negligible friction, the mechanical energy is conserved:

Here h is the height of the slide and g is the acceleration due to gravity. The mass cancels. For example, for a slide of height h = 2 m, v = √(2 × 9.8 × 2) ≈ 6.3 m/s.

Yes (when friction is negligible). A heavier child has more potential energy at the top, but also needs more energy to reach the same speed. Using conservation of energy:

The mass m cancels, so the velocity depends only on the height h of the slide and on g, not on the mass of the child.

Since v = √(2gh), the velocity at the bottom depends only on the vertical height h of the slide. So the slide with the greatest height gives the largest velocity.

The velocity does not depend on the shape, the length or (directly) the steepness of the slide.

Your muscles change chemical energy (from food) into other forms:

1. Kinetic energy: the bicycle and the rider are moving.
2. Heat energy: friction in the chain, the tyres and with the road produces heat, and the body also gives out heat while working.
3. Sound energy: a small amount appears as sound from the tyres and the chain.

1. Equal kinetic energies½ m v_A² = ½ (4m) v_B²
2. Cancel the common termsv_A² = 4 v_B²  ⇒  v_A = 2 v_B

v_A : v_B = 2 : 1

No. The kinetic energy is ½mv². If the velocity is constant, the speed does not change, and the mass is also constant, so the kinetic energy stays the same. Kinetic energy depends on the velocity, not on the position.

• Horizontal motion: No, the potential energy does not change, because it depends on the height (U = mgh) and the height stays the same.
• Raised vertically: Yes, the potential energy increases, because the height h increases.

1. At the start (A, height h)The object is at rest, so KE = 0 and PE = mgh.Total energy = mgh
2. At a point B (height h′)It has fallen a distance (h − h′), so v² = 2g(h − h′).KE = ½mv² = mg(h − h′)    PE = mgh′KE + PE = mg(h − h′) + mgh′ = mgh
3. Just before hitting the ground (C)Here h′ = 0, so PE = 0 and all the energy is kinetic.Mechanical energy = KE = ½mv² = mgh

The mechanical energy just before the ground is mgh.

The ball is released from the highest point, where its potential energy is maximum and its kinetic energy is zero.

• At A (top of the first hill): the potential energy is high and the kinetic energy is smaller, because the ball is slower at a height.
• At B (lowest point in the valley): the potential energy is minimum and the kinetic energy is maximum, because the ball is fastest here.
• At C (top of the next hill): the potential energy is high again, but less than at A, and the kinetic energy is small.

The hills C, D and E are lower than the previous ones because some of the mechanical energy is lost on the way. Yes, it is due to friction and air resistance, which change part of the energy into heat and sound. With less total energy, the ball cannot climb as high.

An inclined ladder acts like an inclined plane. To reach the same height, the same work is done in both cases, but on the inclined ladder it is done over a longer distance, so a smaller force is needed at each step. On a vertical ladder the whole body weight has to be lifted directly upward, which needs more effort.

The spoon works as a lever.

• The rim of the can acts as the fulcrum.
• The spoon handle is a long effort arm, and the tip under the lid is a short load arm.
• Since effort × effort arm = load × load arm, a small force on the long handle produces a large force at the tip, which pops the lid open easily.

Scissors work as a lever with the pivot as the fulcrum. When the object is placed closer to the fulcrum:

• the load arm becomes shorter,
• the effort arm (the handle) becomes relatively longer, and
• the mechanical advantage increases.

So the same effort of the hand produces a larger cutting force on the hard object.

In every real machine, friction and air resistance do negative work on the moving parts. This means:

• the mechanical energy of the machine keeps decreasing,
• the lost energy changes into heat and sound and spreads into the surroundings, and
• no machine can keep its mechanical energy perfectly.

Since the energy is continuously lost and nothing supplies it again, the machine slowly loses its motion and finally stops.

• Lifting up: she applies an upward force equal to the weight, and the dumbbell moves up (force and displacement in the same direction), so she does positive work.
• Lowering down: she still applies an upward force to hold it, but the dumbbell moves down (force and displacement in opposite directions), so she does negative work.

The goalkeeper’s force on the ball acts opposite to the ball’s motion, so the work done is negative.

The work done by the goalkeeper on the ball is −30 J.

The moving striker hits the white coin, which in turn hits the black coin.

• The striker applies a force on the white coin in the direction of its displacement, so the striker does positive work on the white coin and increases its energy. By Newton’s third law, the white coin does negative work on the striker and decreases its energy.
• In the same way, the white coin does positive work on the black coin and increases its energy, while the black coin does negative work on the white coin and decreases its energy.

Original kinetic energy = ½mv². With velocity 2v:

The kinetic energy becomes 4 times the original value.

1. Convert the velocity154.8 km/h = 154.8 × 1000 ÷ 3600 = 43 m/s
2. Kinetic energyKE = ½mv² = ½ × 0.2 × 43² = 184.9 J

KE ≈ 184.9 J

1. Work done by the wireThe force is opposite to the displacement, so the work is negative.W = F × s = 367500 × (−100) = −36,750,000 J
2. Work-energy theoremWork done = change in kinetic energy−36,750,000 = 0 − ½ × 15000 × v²
3. Solve for vv² = 2 × 36,750,000 ÷ 15000 = 4900  ⇒  v = 70 m/s

v = 70 m/s = 252 km/h

The potential energy is 20 J.

1. Initial energyv = 72 km/h = 20 m/sKE = ½ × 10000 × 20² = 2,000,000 J
2. Final energyLet the truck go a distance d up the ramp. At 30° the height gained is d ÷ 2.PE = mg(d/2) = 10000 × 10 × d/2 = 50000 d
3. Work done by the sandW = −50000 d
4. Work-energy theoremWork done = final energy − initial energy−50000 d = 50000 d − 2,000,000  ⇒  100000 d = 2,000,000  ⇒  d = 20 m

The minimum length of the ramp is 20 m.

1. Work doneW = mgh = 75 × 10 × 2 = 1500 J
2. PowerP = W ÷ t = 1500 ÷ 5 = 300 W

Power = 300 W

1. Final speed72 km/h = 20 m/s
2. Work done by the engineWork = change in kinetic energyW = ½ × 1000 × 20² − 0 = 200,000 J
3. PowerP = 200,000 ÷ 10 = 20,000 W

Power = 20,000 W (20 kW)

1. Length of the rampThe height is 30 cm and the base is 40 cm, so by the right-angled triangle property:L = √(30² + 40²) = 50 cm
2. Mechanical advantageMA = load ÷ effort = L ÷ hMA = 50 ÷ 30 ≈ 1.67

The mechanical advantage of the ramp is about 1.67.

For balance, effort × effort arm = load × load arm. Suppose the 15 kg child sits at A (2 m from the fulcrum) and the 30 kg child sits at a distance L on the other side:

The 15 kg child sits at A and the 30 kg child sits at D (or the 15 kg child at E and the 30 kg child at B).

Correct answer: False

Work is done only when the force causes a displacement. If the object does not move, no work is done.

Correct answer: True

The force and the displacement are in the same direction, so the work done is positive.

Correct answer: True

Work and energy are measured in the same unit, the joule.

Correct answer: False

A motionless stretched rubber band has potential energy (elastic potential energy), not kinetic energy.

Correct answer: True

Energy can be transformed from one form to another, but the total energy remains the same.

Answer: Force × Displacement

Answer: 1 newton

Answer: ½mv²

Answer: mgh

Answer: rate

Correct option: (b) Both iii and iv

• (iii) is correct: at the highest point the velocity is momentarily zero, so the kinetic energy is zero.
• (iv) is correct: the height is maximum there, so the potential energy is maximum.
• (i) and (ii) are wrong: gravity still acts on the ball (its weight), so the force and the acceleration (g) are not zero.

1. (i) Lifted straight upPE = mgh = 50 × 10 × 72.5 = 36,250 J
2. (ii) Climbing the stairsThe vertical height gained is the same.PE = mgh = 50 × 10 × 72.5 = 36,250 J
3. (iii) ConclusionThe gain in potential energy is the same in both cases. The potential energy depends only on the initial and final heights, not on the path taken.

(i) 36,250 J    (ii) 36,250 J

1. EnergyLet the height of one floor be h.E₁ = mg(10h) = 10mgh    E₂ = mg(20h) = 20mgh = 2E₁
2. PowerLet the time for the 10th floor be t, so the time for the 20th floor is 2t.P₁ = 10mgh ÷ t    P₂ = 20mgh ÷ 2t = 10mgh ÷ t = P₁

The energy required is double, but the power required is the same.

• Factors: the energy required depends on the mass of the flag and the height to which it is raised (E = mgh).
• Slowly or quickly: the work done does not change, because the same weight is raised through the same height.
• Speed doubled: the power doubles, because the same work is done in half the time (P = W ÷ t).

1. Fuel used is proportional to the kinetic energy gainedKE = ½mv²
2. Day 1Total mass = 60 + 100 = 160 kgKE₁ = ½ × 160 × v²
3. Day 2Total mass = 60 + 40 + 100 = 200 kgKE₂ = ½ × 200 × v²
4. RatioFuel₁ : Fuel₂ = 160 : 200 = 4 : 5

The ratio of fuel used (Day 1 : Day 2) = 4 : 5

Let the weight of the child be W, so the adult weighs 2W. For balance, the clockwise moment equals the anticlockwise moment. If the adult sits at a distance d from the fulcrum and the child at a distance 2d on the other side:

Both moments are equal, so the seesaw is balanced. The child sits twice as far from the fulcrum as the adult.

(i) While the ball goes up, the displacement is upward but gravity acts downward, so the work done by gravity is negative. While it comes down, the displacement and gravity are both downward, so the work done by gravity is positive.

1. (ii) Initial kinetic energyKE = ½ × 2 × 20² = 400 J
2. Potential energy at 19.4 mPE = mgh = 2 × 10 × 19.4 = 388 J
3. Energy lost to air resistance400 − 388 = 12 J

The work done by air resistance is −12 J (12 J of energy is lost).

1. (i) Speed at 0 m180 = ½ × 10 × v²  ⇒  v² = 36  ⇒  v = 6 m/s
2. (ii) Work done from 0 to 4 mThe work is the area under the force-displacement graph, a trapezium with parallel sides 4 m and 2 m and height 50 N.W = ½ × (4 + 2) × 50 = 150 J
3. Kinetic energy at 4 mKE = 180 + 150 = 330 J
4. Speed at 4 m½ × 10 × v² = 330  ⇒  v = √66 ≈ 8.1 m/s

(i) 6 m/s    (ii) ≈ 8.1 m/s

Negative acceleration? No. The force always acts in the direction of motion (it never becomes negative), so the block never slows down. Between 3 m and 4 m the force decreases, so the acceleration decreases, but it stays positive.

1. Height reachedh = v² ÷ 2g
2. On the Earthh_earth = v² ÷ 2g = 8 m
3. On the Moong_moon = g ÷ 6h_moon = v² ÷ 2(g/6) = 6 × h_earth = 6 × 8 = 48 m

The ball will travel 48 m up on the Moon.

1. (i) A to BThe graph is a horizontal line at 35 m/s from 0 to 1 s. The car moves with constant speed (uniform motion). This is the 1 s during which the driver reacts before applying the brakes.
2. (ii) Kinetic energy at AKE = ½ × 1000 × 35² = 500 × 1225 = 612,500 J
3. (iii) Work done by the brakesAt B the speed is 35 m/s and at C it is 0. By the work-energy theorem, work = change in KE.W = 0 − 612,500 = −612,500 JThe work done is negative because the braking force opposes the motion.
4. (iv) Transformation of energyThe kinetic energy changes into heat energy (friction between the brake pads and discs and between the tyres and the road) and a little sound energy (the screech of the brakes).

1. Total mechanical energyAt O the ball is at rest, so the kinetic energy is 0.Total energy = 0 + 30 = 30 J (constant, as there is no friction)
2. At P (PE = 20 J)KE = 30 − 20 = 10 J  ⇒  ½ × 0.5 × v² = 10  ⇒  v = √40 ≈ 6.3 m/s
3. At Q (PE = 30 J)KE = 30 − 30 = 0  ⇒  v = 0
4. At R (PE = 40 J)The kinetic energy would be 30 − 40 = −10 J, which is not possible. The ball cannot reach R; it stops at Q and turns back.

P: ≈ 6.3 m/s    Q: 0    R: the ball cannot reach this point

1. (i) Velocity before impactBy conservation of energy, mgh = ½mv²v = √(2gh) = √(2 × 10 × 10) = √200 ≈ 14.1 m/s
2. (ii) Depth of the depressionEnergy of the coconut = mgh = 1.5 × 10 × 10 = 150 JWork done by the sand = F × d = 3000 × d = 150  ⇒  d = 0.05 m

(i) ≈ 14.1 m/s    (ii) 0.05 m (5 cm)