Chapter 5: Work, Energy and Power Application Questions | physics Class 11 CBSE

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Chapter 5: Work, Energy and Power Application Questions | physics Class 11 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A cyclist applies brakes and comes to rest in 10 m. The retarding force from the road is 200 N. With reference to the work-energy theorem, explain (a) the work done by the road on the cycle and (b) the work done by the cycle on the road.

The stopping (frictional) force and displacement are in opposite directions (θ = 180°), so work done by the road on the cycle is W = 200 × 10 × cos 180° = −2000 J; this negative work removes kinetic energy from the cycle and brings it to rest.
By the work-energy theorem, ΔK = W = −2000 J, which confirms that the cycle's kinetic energy decreases by exactly 2000 J, consistent with the cycle decelerating to rest.
By Newton's third law, the cycle exerts an equal and opposite force (200 N) on the road; however, the road undergoes zero displacement, so the work done by the cycle on the road is zero.
This illustrates that Newton's third law forces are equal and opposite, but the work done by mutual forces on each body need not be equal and opposite; work depends on both force and displacement.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

A woman pushes a trunk on a rough railway platform, applying 100 N over the first 10 m and then a linearly decreasing force from 100 N to 50 N over the next 10 m. The frictional force is 50 N throughout. Calculate the net work done on the trunk over 20 m.

Work done by the woman in the first 10 m (constant force 100 N): W1 = 100 × 10 = 1000 J; work done in the next 10 m (trapezoidal area): W2 = (1/2)(100 + 50) × 10 = 750 J; total work by woman WF = 1750 J.
The frictional force of 50 N acts opposite to displacement over the full 20 m: Wf = −50 × 20 = −1000 J; the negative sign indicates friction opposes motion throughout.
Net work done on the trunk = WF + Wf = 1750 + (−1000) = 750 J; by the work-energy theorem, the kinetic energy of the trunk increases by 750 J over the 20 m displacement.
This problem illustrates that for a variable applied force, work is computed as the area under the F-x graph, while friction's constant-force negative work is calculated directly using W = Fd cos 180°.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

A bullet of mass 50 g is fired at 200 m/s into soft plywood and emerges with only 10% of its initial kinetic energy. Using the work-energy theorem, find the emergent speed of the bullet.

Initial kinetic energy of the bullet: Ki = (1/2) × 0.05 × 200² = (1/2) × 0.05 × 40000 = 1000 J.
Final kinetic energy Kf = 10% of Ki = 0.1 × 1000 = 100 J; the bullet loses 900 J of kinetic energy to work done against the plywood.
Using Kf = (1/2)mvf²: 100 = (1/2) × 0.05 × vf², so vf² = 100 × 2 / 0.05 = 4000, giving vf = 63.2 m/s.
Although the bullet loses 90% of its kinetic energy, its speed is reduced by only about 68% (not 90%), because kinetic energy is proportional to v² — a result that highlights the non-linear relationship between speed and kinetic energy.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

A car of mass 1000 kg moving at 18 km/h collides with a spring of spring constant 5.25 × 10³ N/m on a smooth surface. Find the maximum compression of the spring.

Convert speed: 18 km/h = 5 m/s; initial kinetic energy of car: K = (1/2) × 1000 × 5² = (1/2) × 1000 × 25 = 12500 J = 1.25 × 10⁴ J.
At maximum compression xm, the car is momentarily at rest; by conservation of mechanical energy, all kinetic energy converts to spring potential energy: (1/2)kxm² = K.
Solving for xm: xm² = 2K/k = 2 × 1.25 × 10⁴ / 5.25 × 10³ = 25000/5250 ≈ 4.76 m²; xm = √4.76 ≈ 2.00 m.
This idealised calculation assumes the spring is massless and the surface is frictionless; introducing friction (μ = 0.5) reduces the compression to 1.35 m, showing that non-conservative forces significantly reduce the energy available to compress the spring.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

An elevator of total mass 1800 kg moves upward at a constant speed of 2 m/s against a frictional force of 4000 N. Determine the minimum power the motor must deliver in watts and horsepower.

Since the elevator moves at constant speed, the net force on it is zero; the motor must supply a force equal to the total downward force: F = mg + Ff = (1800 × 10) + 4000 = 22000 N.
Minimum power delivered by the motor: P = F·v = 22000 × 2 = 44000 W = 44 kW.
Converting to horsepower: P = 44000 / 746 ≈ 59 hp; this is the minimum power required to maintain constant upward motion.
At constant velocity, all power input goes into overcoming gravity (useful work) and friction (dissipated as heat); if the speed were higher, more power would be required since P = F·v increases with v.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

With reference to the concept of potential energy, explain why fault lines in the earth's crust can cause earthquakes, and how this is analogous to a compressed spring.

The earth's crust has discontinuities called fault lines which experience enormous compressive tectonic forces over geological timescales; work done by these forces gets stored as elastic potential energy in the compressed rock.
This is analogous to a compressed spring (Fs = −kx), where work done against the spring force is stored as V = kx²/2; the fault lines behave like very stiff, large-scale compressed springs.
When the accumulated potential energy exceeds the strength of the rock, the fault line suddenly readjusts (slips); the stored potential energy converts rapidly to kinetic energy of the surrounding rock, generating seismic waves — an earthquake.
The chapter notes that potential energy is the 'stored energy by virtue of position or configuration'; fault lines embody this definition on a massive scale, with catastrophic release when constraints are removed.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 7

Applied Concept

A pump on the ground floor of a building pumps water to fill a 30 m³ tank located 40 m above the ground in 15 minutes. If the pump efficiency is 30%, find the electric power consumed.

Mass of water: m = volume × density = 30 m³ × 1000 kg/m³ = 3 × 10⁴ kg; work done against gravity: W = mgh = 3 × 10⁴ × 10 × 40 = 1.2 × 10⁷ J.
Time taken: t = 15 min = 900 s; useful (output) power of pump: Puseful = W/t = 1.2 × 10⁷ / 900 ≈ 1.333 × 10⁴ W.
Since pump efficiency = useful power / input power = 30% = 0.30, the electric power consumed: Pin = Puseful / 0.30 = 1.333 × 10⁴ / 0.30 ≈ 4.4 × 10⁴ W = 44 kW.
The efficiency of 30% means that 70% of the electrical energy input is wasted as heat in the motor and mechanical friction, and only 30% goes into lifting the water; improving pump efficiency reduces electricity bills significantly.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 8

Applied Concept

In a nuclear reactor, a neutron collides elastically with a deuterium nucleus (mass = 2 × neutron mass). What fraction of the neutron's kinetic energy is transferred to the deuterium nucleus? Why is this important for reactor operation?

For an elastic collision with neutron mass m1 and deuterium mass m2 = 2m1: the fractional energy retained by the neutron is f1 = [(m1−2m1)/(m1+2m1)]² = (−1/3)² = 1/9 ≈ 11%.
The fractional energy transferred to deuterium: f2 = 1 − f1 = 1 − 1/9 = 8/9 ≈ 89%; nearly 90% of the neutron's kinetic energy is given to deuterium in a single head-on collision.
Fast neutrons (≈10⁷ m/s) from fission must be slowed to ≈10³ m/s to efficiently cause further fissions in U-235; heavy water (D₂O, containing deuterium) is an effective moderator because of this high energy transfer fraction.
Carbon (graphite) is a less efficient moderator: f2 ≈ 28.4% per collision; thus more collisions are needed with graphite to slow neutrons to thermal speeds, but it is cheaper and easier to handle than heavy water.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 9

Applied Concept

A bob of mass m is released from a horizontal position and swings to the lowest point of a pendulum of length 1.5 m, dissipating 5% of its initial energy against air resistance. Find its speed at the lowest point.

Initial potential energy (taking lowest point as reference): V = mgL = m × 10 × 1.5 = 15m J; this is the total mechanical energy at the starting horizontal position (K = 0).
Energy dissipated by air resistance = 5% of initial energy = 0.05 × 15m = 0.75m J; energy remaining at the bottom = 15m − 0.75m = 14.25m J.
At the lowest point, all remaining energy is kinetic: (1/2)mv² = 14.25m, so v² = 28.5 m²/s² and v = √28.5 ≈ 5.34 m/s.
Without air resistance (100% energy), v = √(2gL) = √30 ≈ 5.48 m/s; the 5% energy loss reduces the speed by less than 3%, showing that air resistance has a relatively small effect on the final speed.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

A trolley (300 kg) carrying a sandbag (25 kg) moves at 27 km/h on a frictionless track. Sand leaks out at 0.05 kg/s until the bag is empty. Find the speed of the trolley after all sand has leaked out.

The leaking sand exerts no horizontal force on the trolley as it falls vertically; therefore no horizontal external force acts on the trolley-sandbag system in the horizontal direction.
By conservation of linear momentum in the horizontal direction, the total momentum is conserved; the leaking sand carries away zero horizontal momentum since it falls vertically.
Initial momentum = (300 + 25) × 27 km/h = 325 × 7.5 m/s = 2437.5 kg·m/s; final mass of trolley (sand gone) = 300 kg; by conservation of momentum, 300 × v = 2437.5, so v = 8.125 m/s = 29.25 km/h.
The trolley actually speeds up slightly when the sandbag empties on a frictionless track, because the momentum is now shared among fewer kilograms; the sand carries away no horizontal momentum.

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Chapter 5: Work, Energy and Power | Application Questions

A cyclist applies brakes and comes to rest in 10 m. The retarding force from the road is 200 N. With reference to the work-energy theorem, explain (a) the work done by the road on the cycle and (b) the work done by the cycle on the road.

A woman pushes a trunk on a rough railway platform, applying 100 N over the first 10 m and then a linearly decreasing force from 100 N to 50 N over the next 10 m. The frictional force is 50 N throughout. Calculate the net work done on the trunk over 20 m.

A bullet of mass 50 g is fired at 200 m/s into soft plywood and emerges with only 10% of its initial kinetic energy. Using the work-energy theorem, find the emergent speed of the bullet.

A car of mass 1000 kg moving at 18 km/h collides with a spring of spring constant 5.25 × 10³ N/m on a smooth surface. Find the maximum compression of the spring.

An elevator of total mass 1800 kg moves upward at a constant speed of 2 m/s against a frictional force of 4000 N. Determine the minimum power the motor must deliver in watts and horsepower.

With reference to the concept of potential energy, explain why fault lines in the earth's crust can cause earthquakes, and how this is analogous to a compressed spring.

A pump on the ground floor of a building pumps water to fill a 30 m³ tank located 40 m above the ground in 15 minutes. If the pump efficiency is 30%, find the electric power consumed.

In a nuclear reactor, a neutron collides elastically with a deuterium nucleus (mass = 2 × neutron mass). What fraction of the neutron's kinetic energy is transferred to the deuterium nucleus? Why is this important for reactor operation?

A bob of mass m is released from a horizontal position and swings to the lowest point of a pendulum of length 1.5 m, dissipating 5% of its initial energy against air resistance. Find its speed at the lowest point.

A trolley (300 kg) carrying a sandbag (25 kg) moves at 27 km/h on a frictionless track. Sand leaks out at 0.05 kg/s until the bag is empty. Find the speed of the trolley after all sand has leaked out.

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