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

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

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Rohit, a Class 11 student, performs a lab experiment with a spring-block system on a smooth horizontal surface. He attaches a block of mass 2 kg to a spring of spring constant 400 N/m. He pulls the block to a displacement of 0.1 m from the equilibrium position and releases it from rest. He observes that the block oscillates back and forth. He notes the kinetic and potential energies at different positions and tries to verify the conservation of mechanical energy. He also calculates the maximum speed of the block at the equilibrium position and plots the energy-displacement graph to see how kinetic and potential energies vary as complementary parabolic curves.

Question 1: What is the total mechanical energy of the spring-block system when the block is at maximum displacement xm = 0.1 m?

At maximum displacement, the block is momentarily at rest so kinetic energy K = 0; all energy is stored as elastic potential energy in the spring.
Total mechanical energy E = V(xm) = (1/2)kxm² = (1/2) × 400 × (0.1)² = (1/2) × 400 × 0.01 = 2 J.
This total energy E = 2 J remains constant throughout the oscillation since the spring force is conservative and the surface is smooth (no friction).

Question 2: Find the maximum speed of the block. At which position does it occur and why?

Maximum speed occurs at the equilibrium position x = 0, where potential energy V = 0 and all mechanical energy is kinetic.
Setting (1/2)mv²_max = E = 2 J: v²_max = 2E/m = 2 × 2 / 2 = 2 m²/s², so v_max = √2 ≈ 1.41 m/s.
At equilibrium, the spring force is zero (Fs = −kx = 0), so no net force acts to decelerate the block; beyond this point the restoring force decelerates it, making the equilibrium the point of maximum speed.

Question 3: If a frictional surface (μ = 0.1) is now used instead of the smooth surface, use the work-energy theorem to find the maximum compression when the same block is pushed against the spring with initial KE of 2 J. (g = 10 m/s²)

With friction, both spring force and friction oppose the compression; applying the work-energy theorem: ΔK = W_spring + W_friction, so 0 − 2 = −(1/2)kxm² − μmgxm.
This gives (1/2)(400)xm² + (0.1)(2)(10)xm − 2 = 0, i.e., 200xm² + 2xm − 2 = 0, or 100xm² + xm − 1 = 0.
Using the quadratic formula: xm = [−1 + √(1 + 400)] / 200 = [−1 + √401] / 200 ≈ [−1 + 20.02] / 200 ≈ 19.02/200 ≈ 0.095 m; the compression is less than 0.1 m (frictionless case) as expected, since friction dissipates some energy.

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Case Set 2

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During a physics demonstration, a teacher suspends a bob of mass 0.5 kg by a light inextensible string of length 2 m. She pulls the bob to one side so that the string makes 60° with the vertical and releases it from rest. The bob swings down through the lowest point and continues upward on the other side. The teacher asks students to track the energy transformations at various points during the swing and to find the speed at the lowest point. She also asks them to consider what would happen if 10% of the mechanical energy is lost to air resistance during the descent from the initial position to the lowest point.

Question 1: Taking the lowest point as the reference level, what is the initial potential energy of the bob when the string makes 60° with the vertical?

The height of the bob above the lowest point: h = L − L cos 60° = L(1 − cos 60°) = 2(1 − 0.5) = 2 × 0.5 = 1 m.
Initial potential energy V = mgh = 0.5 × 10 × 1 = 5 J; at this position the bob is at rest so kinetic energy K = 0 and total mechanical energy E = 5 J.
This potential energy is entirely due to the gravitational force; the string tension does no work since displacement is always perpendicular to the string.

Question 2: Find the speed of the bob at the lowest point assuming no energy loss (smooth swing).

At the lowest point, all potential energy converts to kinetic energy (reference level); by conservation of mechanical energy: (1/2)mv² = mgh = 5 J.
Solving: v² = 2 × 5 / 0.5 = 20 m²/s², so v = √20 = 2√5 ≈ 4.47 m/s.
This result is independent of the mass of the bob and depends only on the height h = 1 m and g = 10 m/s²; v = √(2gh) = √(2 × 10 × 1) = √20 m/s.

Question 3: If 10% of the mechanical energy is dissipated by air resistance during the swing from the initial position to the lowest point, find the new speed at the lowest point and the work done by air resistance.

Energy available at lowest point = 90% of initial energy = 0.9 × 5 = 4.5 J; all this is kinetic at the lowest point: (1/2)(0.5)v² = 4.5.
Solving: v² = 4.5 × 2 / 0.5 = 18 m²/s², so v = √18 = 3√2 ≈ 4.24 m/s; compared to 4.47 m/s (no loss), the speed is reduced by about 5%.
Work done by air resistance = −10% of E = −0.5 J (negative because it opposes motion); this confirms Ef − Ei = Wnc: 4.5 − 5 = −0.5 J, which is the work done by the non-conservative air resistance force.

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Case Set 3

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In a nuclear power plant, a technician explains to a group of students how fast neutrons released during fission of uranium-235 must be slowed down to 'thermal' speeds before they can efficiently cause further fissions. The plant uses heavy water (D₂O) as a moderator. Each deuterium nucleus (mass ≈ 2 × neutron mass) can slow a neutron significantly in a single elastic collision. The technician notes that graphite (carbon, mass ≈ 12 × neutron mass) is used in some reactors but is less efficient per collision. Students are asked to compare the two moderators using the fractional kinetic energy transfer formula derived from elastic collision theory.

Question 1: State the conditions that must be satisfied during the collision between a neutron and a moderator nucleus for the elastic collision formulae to apply.

The collision must be elastic, meaning both total linear momentum and total kinetic energy must be conserved before and after the collision.
The moderator nucleus must be initially at rest (stationary target); no loss of generality is involved in this assumption as stated in the chapter.
The collision is treated as a one-dimensional (head-on) interaction for the purpose of calculating maximum energy transfer; the two bodies are treated as point masses with no rotational effects.

Question 2: Calculate the fraction of the neutron's kinetic energy transferred to the deuterium nucleus (m₂ = 2m₁) in a single head-on elastic collision.

For elastic collision: final velocity of moderator nucleus v2f = 2m1/(m1+m2) × v1i; for deuterium m2 = 2m1: v2f = 2m1/(3m1) × v1i = (2/3)v1i.
Kinetic energy of deuterium after collision: K2f = (1/2)(2m1)(2v1i/3)² = (1/2)(2m1)(4v1i²/9) = (4m1v1i²/9).
Fractional energy transferred: f2 = K2f/K1i = (4m1v1i²/9) / ((1/2)m1v1i²) = 8/9 ≈ 89%; nearly 90% of the neutron's kinetic energy is transferred to deuterium in one head-on collision.

Question 3: Compare the efficiency of deuterium and carbon as moderators using the fractional energy retained by the neutron, and explain why heavy water is preferred despite being more expensive.

For deuterium (m2 = 2m1): f1 = [(m1−2m1)/(m1+2m1)]² = (1/9) ≈ 11%; the neutron retains only 11% of its energy, transferring 89% per collision — extremely efficient.
For carbon (m2 = 12m1): f1 = [(m1−12m1)/(m1+12m1)]² = (11/13)² ≈ 71.6%; the neutron retains about 72% of its energy, transferring only 28.4% per collision — much less efficient.
Therefore, fewer collisions are needed with deuterium to slow a neutron from ~10⁷ m/s to ~10³ m/s; despite the higher cost of heavy water, it is preferred in CANDU reactors for its superior moderation efficiency and because it also allows use of natural (unenriched) uranium as fuel.

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Case Set 4

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An automobile company tests the safety of its cars by simulating collisions with a spring-mounted barrier. In one test, a car of mass 1200 kg moving at 36 km/h on a smooth road collides with a horizontally mounted spring of spring constant 6 × 10³ N/m. In a second test, the same car at the same speed is tested on a road with coefficient of kinetic friction 0.4. The engineers record the maximum compression in both cases and compare them to study the effect of road surface on collision safety. They note that the spring is massless and the barrier does not move.

Question 1: Convert the car's speed and find its initial kinetic energy before the collision.

Speed conversion: 36 km/h = 36 × (1000/3600) = 10 m/s; this is a useful conversion — 36 km/h = 10 m/s.
Initial kinetic energy: K = (1/2)mv² = (1/2) × 1200 × 10² = (1/2) × 1200 × 100 = 60000 J = 6 × 10⁴ J.
This kinetic energy must be fully absorbed either by the spring (frictionless case) or shared between the spring and friction (rough road case) at maximum compression.

Question 2: Find the maximum compression of the spring on the smooth road using conservation of mechanical energy.

On the smooth road, at maximum compression xm the car is at rest; by conservation of mechanical energy: K = V_spring, so (1/2)kxm² = 6 × 10⁴ J.
Solving: xm² = 2 × 6×10⁴ / 6×10³ = 120000/6000 = 20 m²; xm = √20 = 2√5 ≈ 2.24 m (approximately 2.0 m for rounded values).
This calculation assumes a massless spring and a frictionless surface; in reality, energy is also stored in the car's deformation, making actual compressions smaller.

Question 3: On the rough road (μ = 0.4, g = 10 m/s²), set up the work-energy equation and find the maximum compression of the spring. Compare with the smooth road result.

On the rough road, both spring force and friction oppose compression; by the work-energy theorem: 0 − K = −(1/2)kxm² − μmgxm, giving (1/2)(6000)xm² + (0.4)(1200)(10)xm − 60000 = 0.
Simplifying: 3000xm² + 4800xm − 60000 = 0, or xm² + 1.6xm − 20 = 0; using the quadratic formula: xm = [−1.6 + √(2.56 + 80)] / 2 = [−1.6 + √82.56] / 2 ≈ [−1.6 + 9.086] / 2 ≈ 3.74 m.
Wait — taking positive root: xm ≈ (−1.6 + 9.09)/2 ≈ 3.74 m seems larger; re-checking: 3000xm² + 4800xm − 60000 = 0 → xm² + 1.6xm − 20 = 0 → xm = [−1.6 + √(2.56+80)]/2 ≈ 3.74 m is incorrect since friction should reduce compression. Correct setup: ΔK = W_net: −60000 = −3000xm² − 4800xm → 3000xm² + 4800xm − 60000 = 0 → xm ≈ 3.74 m is mathematically correct but physically means friction on the return path is absent in this one-way compression analysis; the compression is less intuitive — friction reduces available energy to compress spring but the quadratic naturally gives the correct positive root ≈ 3.74 m is wrong. Correct: xm = [−1.6+√(1.6²+4×20)]/2 = [−1.6+9.09]/2 ≈ 3.74 but this exceeds frictionless; error is in sign — friction helps decelerate: 60000 = 3000xm²+4800xm → xm²+1.6xm−20=0 → xm≈3.74. Since 3000(3.74)²+4800(3.74)=3000(13.99)+17952=41970+17952=59922≈60000 ✓. Friction on smooth gave xm=√20≈4.47 m; rough road gives 3.74 m — friction reduces compression, confirming the result is physically correct (rough surface dissipates energy so less is available to compress spring, giving smaller compression).

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Case Set 5

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During a science fair, a student demonstrates the conversion of wind energy into electrical energy using a small model windmill. The windmill blades sweep a circular area of radius 0.5 m. On the day of the demonstration, wind blows at 36 km/h perpendicular to the blade circle. The density of air is 1.2 kg/m³. The model windmill converts 25% of the wind's kinetic energy into electrical energy. The student measures the electrical power output and explains to visitors how the same principle is used in large wind farms, where blade radii can be 50 m or more and wind speeds can be much higher, leading to megawatt-scale power outputs.

Question 1: Find the mass of air flowing through the blade circle per second.

Area swept by blades: A = πr² = π × (0.5)² = π/4 ≈ 0.785 m².
Wind speed: v = 36 km/h = 10 m/s; volume of air flowing through area A per second = A × v = 0.785 × 10 = 7.85 m³/s.
Mass of air per second: dm/dt = ρ × Av = 1.2 × 7.85 ≈ 9.42 kg/s; this is the mass flow rate of air that passes through the windmill's sweep area each second.

Question 2: Calculate the total wind power available and the electrical power produced by the model windmill.

Kinetic energy of air per second (wind power) = (1/2)(dm/dt)v² = (1/2)(9.42)(10²) = (1/2)(9.42)(100) = 471 W.
Electrical power produced = 25% of wind power = 0.25 × 471 ≈ 117.75 W ≈ 118 W.
This is the power output from a small model with 0.5 m blades; real wind turbines with 50 m blades (area 7854 m²) at the same wind speed would produce about 10,000 times more power — illustrating how blade area critically determines power output.

Question 3: If the wind speed doubles to 72 km/h, by what factor does the electrical power output increase? Explain the physical reason for this relationship.

New wind speed v' = 72 km/h = 20 m/s = 2v; new mass flow rate dm'/dt = ρAv' = 1.2 × 0.785 × 20 = 18.84 kg/s = 2 × (dm/dt).
New wind power = (1/2)(dm'/dt)(v')² = (1/2)(2 × dm/dt)(2v)² = (1/2)(2)(4)(dm/dt)v² = 8 × (original wind power).
The electrical power increases by a factor of 8 (cube of speed ratio = 2³); wind power depends on mass flow rate (∝ v) multiplied by v² from kinetic energy, giving overall v³ dependence — this cubic relationship means small increases in wind speed lead to dramatic increases in power output, which is why wind farms are sited in high-wind locations.

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Chapter 5: Work, Energy and Power | Case Study Questions

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Question 1: What is the total mechanical energy of the spring-block system when the block is at maximum displacement xm = 0.1 m?

Question 2: Find the maximum speed of the block. At which position does it occur and why?

Question 3: If a frictional surface (μ = 0.1) is now used instead of the smooth surface, use the work-energy theorem to find the maximum compression when the same block is pushed against the spring with initial KE of 2 J. (g = 10 m/s²)

Passage

Question 1: Taking the lowest point as the reference level, what is the initial potential energy of the bob when the string makes 60° with the vertical?

Question 2: Find the speed of the bob at the lowest point assuming no energy loss (smooth swing).

Question 3: If 10% of the mechanical energy is dissipated by air resistance during the swing from the initial position to the lowest point, find the new speed at the lowest point and the work done by air resistance.

Passage

Question 1: State the conditions that must be satisfied during the collision between a neutron and a moderator nucleus for the elastic collision formulae to apply.

Question 2: Calculate the fraction of the neutron's kinetic energy transferred to the deuterium nucleus (m₂ = 2m₁) in a single head-on elastic collision.

Question 3: Compare the efficiency of deuterium and carbon as moderators using the fractional energy retained by the neutron, and explain why heavy water is preferred despite being more expensive.

Passage

Question 1: Convert the car's speed and find its initial kinetic energy before the collision.

Question 2: Find the maximum compression of the spring on the smooth road using conservation of mechanical energy.

Question 3: On the rough road (μ = 0.4, g = 10 m/s²), set up the work-energy equation and find the maximum compression of the spring. Compare with the smooth road result.

Passage

Question 1: Find the mass of air flowing through the blade circle per second.

Question 2: Calculate the total wind power available and the electrical power produced by the model windmill.

Question 3: If the wind speed doubles to 72 km/h, by what factor does the electrical power output increase? Explain the physical reason for this relationship.

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