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

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

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Question 1

Long Form

Derive the work-energy theorem for a variable force in one dimension and explain its two key limitations compared to Newton's second law.

The time rate of change of kinetic energy is dK/dt = d(mv²/2)/dt = mv(dv/dt) = Fv = F(dx/dt), where Newton's second law F = m(dv/dt) has been used.
This gives dK = F dx; integrating from initial position xi to final position xf yields ∫dK = ∫F dx, so Kf − Ki = W, where W is the work done by the variable force.
This result proves the work-energy theorem for a variable force: the change in kinetic energy equals the work done by the net variable force, generalising the constant-force result.
First limitation: temporal information is lost — the WE theorem results from integrating Newton's second law over time, so it does not reveal how the motion evolves at each instant.
Second limitation: directional information is absent — Newton's second law is a vector equation valid in 3D, while the WE theorem is scalar, so information about the direction of motion is not retained.
Despite these limitations, the WE theorem is powerful for finding speeds at specific positions without needing the complete force-time history, making it ideal for variable-force problems.

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Question 2

Long Form

Derive the expression for the potential energy of a spring and describe graphically how kinetic and potential energy vary with displacement in a spring-block system.

For an ideal spring with spring constant k, Hooke's law gives the spring force as Fs = −kx; the negative sign indicates the restoring nature of the force opposing displacement x from equilibrium.
Work done by the spring when the block is pulled to extension xm is Ws = −∫kx dx from 0 to xm = −kxm²/2; this work is negative because force opposes displacement.
The potential energy is defined as the negative of work done by the spring force: V(x) = kx²/2; this function is always non-negative and parabolic in x.
By conservation of mechanical energy, the total energy E = (1/2)mv² + (1/2)kx² = (1/2)kxm² (constant); kinetic energy is maximum (= E) at x = 0 and zero at x = ±xm.
On the energy-displacement graph (Fig. 5.8), V(x) = kx²/2 is an upward parabola and K(x) = E − kx²/2 is a downward parabola; the two are complementary — when one increases the other decreases by the same amount.
The total mechanical energy E = K + V remains constant as a horizontal line on the graph, confirming conservation of mechanical energy for this conservative spring force.

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Question 3

Long Form

Derive expressions for the final velocities of two bodies in a one-dimensional elastic collision and discuss the two special cases that arise from the result.

For a one-dimensional elastic collision with m1 moving at v1i and m2 at rest, conservation of momentum gives m1v1i = m1v1f + m2v2f (Eq. 5.23) and conservation of KE gives m1v1i² = m1v1f² + m2v2f² (Eq. 5.24).
Rearranging these equations: m1(v1i² − v1f²) = m2v2f², and m1(v1i − v1f) = m2v2f; dividing gives v1i + v1f = v2f, meaning the relative velocity of approach equals relative velocity of separation.
Substituting back, the final velocities are v1f = (m1 − m2)/(m1 + m2) × v1i and v2f = 2m1/(m1 + m2) × v1i.
Special Case I (equal masses, m1 = m2): v1f = 0 and v2f = v1i — the first mass comes to rest and the second moves with the first mass's initial speed; complete transfer of momentum and energy occurs.
Special Case II (m2 >> m1): v1f ≈ −v1i and v2f ≈ 0 — the lighter mass bounces back with nearly the same speed while the heavier mass remains essentially undisturbed.
These results have practical significance: Case I explains billiard ball behaviour when equal masses collide head-on, and Case II explains why a ball bounces back from a heavy wall.

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Question 4

Long Form

Explain the concept of conservation of mechanical energy with the example of a ball dropped from a height H. Verify the conservation at three different heights.

A ball of mass m dropped from height H has purely potential energy at the top: EH = mgH, with no kinetic energy since the initial velocity is zero.
At an intermediate height h (0 < h < H), the ball has fallen through (H − h), gaining speed vh² = 2g(H − h); the total energy is Eh = mgh + (1/2)mvh² = mgh + mg(H − h) = mgH.
At the ground (h = 0), all potential energy has converted to kinetic energy: E0 = (1/2)mvf², and since vf² = 2gH, E0 = mgH.
Thus EH = Eh = E0 = mgH, demonstrating that the total mechanical energy is conserved at every point during the fall under gravity.
The individual energies vary: potential energy decreases linearly from mgH to 0 while kinetic energy increases from 0 to mgH, but their sum remains constant at mgH.
This holds because gravity is a conservative force; if non-conservative forces like air resistance were present, mechanical energy would decrease, with the lost energy converted into thermal energy.

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Question 5

Long Form

Analyse the motion of a bob suspended by a string of length L given a horizontal velocity v0 at the bottom such that it just completes a vertical circle, obtaining the minimum v0 and the ratio of kinetic energies at the midpoint B and topmost point C.

Two forces act on the bob: gravity (conservative) and string tension (does no work as it is always perpendicular to displacement); so total mechanical energy is conserved throughout the circular motion.
At the topmost point C, the string just becomes slack when tension Tc = 0; applying Newton's second law for circular motion: mg = mvc²/L, giving vC = √(gL).
Setting total mechanical energy equal at A (bottom) and C (top): (1/2)mv0² = (1/2)mvC² + 2mgL; substituting vC² = gL gives (1/2)mv0² = (5/2)mgL, so v0 = √(5gL).
At the midpoint B (height L from A), energy conservation gives (1/2)mvB² + mgL = (5/2)mgL, so vB² = 3gL and vB = √(3gL).
The ratio of kinetic energies is KB/KC = (1/2)mvB² / (1/2)mvC² = 3gL / gL = 3/1.
After reaching C, if the string is cut the bob undergoes projectile motion with horizontal velocity √(gL) to the left; otherwise it continues circular motion and completes the revolution.

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Question 6

Long Form

Compare elastic and inelastic collisions in one dimension, deriving the loss in kinetic energy for a completely inelastic collision and showing it is always positive.

In a completely inelastic collision, both bodies move together after collision; momentum conservation gives m1v1i = (m1 + m2)vf, so vf = m1v1i/(m1 + m2).
Initial kinetic energy Ki = (1/2)m1v1i²; final kinetic energy Kf = (1/2)(m1+m2)vf² = (1/2)(m1+m2)[m1/(m1+m2)]²v1i² = (1/2)[m1²/(m1+m2)]v1i².
Loss ΔK = Ki − Kf = (1/2)m1v1i²[1 − m1/(m1+m2)] = (1/2)[m1m2/(m1+m2)]v1i², which is always positive since all quantities are positive.
In an elastic collision, both momentum and kinetic energy are conserved; final velocities are v1f = (m1−m2)/(m1+m2)v1i and v2f = 2m1/(m1+m2)v1i, and the net kinetic energy loss is zero.
The completely inelastic case represents maximum kinetic energy loss for given initial conditions; any other collision (inelastic) lies between this extreme and the elastic case.
The lost kinetic energy in inelastic collisions transforms into internal energy (heat, sound, deformation), which is not recoverable as kinetic energy, distinguishing it fundamentally from elastic collisions.

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Question 7

Long Form

Explain three equivalent definitions of a conservative force and show, using the spring force, that it satisfies all three definitions.

Definition 1: A conservative force F(x) can be derived from a scalar potential energy function V(x) by F(x) = −dV/dx. For the spring, V(x) = kx²/2, and −dV/dx = −kx = Fs, confirming this definition is satisfied.
Definition 2: The work done by a conservative force depends only on the end points and not on the path. For the spring, Ws = (kxi²/2) − (kxf²/2), which depends only on xi and xf, not on the path between them.
Definition 3: The work done by a conservative force over any closed path is zero. For the spring, if the block starts at xi and returns to xi, Ws = (kxi²/2) − (kxi²/2) = 0 (Eq. 5.18).
The spring force Fs = −kx explicitly satisfies all three definitions simultaneously, making it a canonical example of a conservative force in mechanics.
By contrast, friction does not satisfy any of these definitions: no potential energy function exists for friction, its work depends on path length, and work done by friction in a closed loop is negative (never zero).
The physical implication is that for conservative forces, energy stored as potential energy is fully recoverable as kinetic energy, enabling the principle of conservation of mechanical energy to hold.

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Question 8

Long Form

Describe the scalar product of two vectors in detail, including the algebraic expression in component form, the laws it obeys, and its use in computing work done.

The scalar (dot) product of vectors A and B is defined as A·B = AB cos θ, where θ is the angle between the vectors; it yields a dimensionless scalar when A and B are unit vectors, or a scalar with appropriate units otherwise.
In component form, A·B = AxBx + AyBy + AzBz; this follows from expanding the dot products of unit vectors, where i·i = j·j = k·k = 1 and i·j = j·k = k·i = 0.
The scalar product obeys: commutative law A·B = B·A; distributive law A·(B+C) = A·B + A·C; and A·(λB) = λ(A·B) for any real scalar λ.
A useful identity is A·A = Ax² + Ay² + Az² = A², giving the magnitude of a vector; also, A·B = 0 if A and B are perpendicular (θ = 90°), regardless of their magnitudes.
Work done by force F over displacement d is W = F·d = Fd cos θ = Fxdx + Fydy + Fzdz; only the component of force along the displacement contributes to work.
For the special case of unit vectors along coordinate axes, this allows efficient calculation of work in three-dimensional problems where force and displacement have multiple components.

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Long Answer

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Hard

Question 9

Long Form

Derive the formula for maximum compression of a spring when a car of mass m collides with it at speed v, and explain how the result changes if friction (coefficient μ) is present on the surface.

Without friction (smooth surface): at maximum compression xm, the car comes to rest and all kinetic energy converts to spring potential energy by conservation of mechanical energy.
Setting (1/2)mv² = (1/2)kxm² and solving, xm = v√(m/k); for m = 1000 kg, v = 5 m/s (18 km/h converted), k = 5.25×10³ N/m, xm = 2.00 m.
With friction (coefficient μ): both the spring force and frictional force (μmg) oppose the compression; applying the work-energy theorem (not just energy conservation), ΔK = Wspring + Wfriction.
The equation becomes 0 − (1/2)mv² = −(1/2)kxm² − μmgxm, rearranging to kxm² + 2μmgxm − mv² = 0, a quadratic in xm.
Solving the quadratic: xm = [−μmg + √(μ²m²g² + mkv²)] / k; for μ = 0.5, m = 1000 kg, this gives xm = 1.35 m, which is less than the frictionless case (2.00 m) as expected.
The reduction in compression with friction shows that non-conservative forces reduce the energy available to compress the spring; unlike the frictionless case, mechanical energy is not conserved here.

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Question 10

Long Form

Explain the concept of power in physics, derive the relation P = F·v, and discuss the practical significance of the units watt and horsepower with examples.

Power is defined as the rate of doing work; average power Pav = W/t; instantaneous power is the limiting value P = dW/dt as time interval approaches zero.
The work done by force F for an infinitesimal displacement dr is dW = F·dr; dividing by dt gives P = F·(dr/dt) = F·v, where v is the instantaneous velocity vector.
Power is a scalar quantity with dimensions [ML²T⁻³]; its SI unit is the watt (W) = 1 J s⁻¹, named after James Watt, an innovator of the steam engine in the eighteenth century.
The horsepower (1 hp = 746 W) is used to rate automobile and motorbike engines; a person physically fit enough to climb four floors quickly is said to be more powerful than one who does it slowly, even if both do the same total work.
For an elevator of total mass 1800 kg moving upward at 2 m/s against a frictional force of 4000 N: total downward force = 18000 + 4000 = 22000 N, so minimum motor power = F·v = 22000 × 2 = 44000 W = 59 hp.
The energy unit kWh = 3.6×10⁶ J appears in electricity bills; a 100 W bulb burning for 10 hours uses 1 kWh; kWh measures energy consumed, while watt and horsepower measure the rate of energy use.

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Chapter sections

Chapter 5: Work, Energy and Power | Long Questions

Derive the work-energy theorem for a variable force in one dimension and explain its two key limitations compared to Newton's second law.

Derive the expression for the potential energy of a spring and describe graphically how kinetic and potential energy vary with displacement in a spring-block system.

Derive expressions for the final velocities of two bodies in a one-dimensional elastic collision and discuss the two special cases that arise from the result.

Explain the concept of conservation of mechanical energy with the example of a ball dropped from a height H. Verify the conservation at three different heights.

Analyse the motion of a bob suspended by a string of length L given a horizontal velocity v0 at the bottom such that it just completes a vertical circle, obtaining the minimum v0 and the ratio of kinetic energies at the midpoint B and topmost point C.

Compare elastic and inelastic collisions in one dimension, deriving the loss in kinetic energy for a completely inelastic collision and showing it is always positive.

Explain three equivalent definitions of a conservative force and show, using the spring force, that it satisfies all three definitions.

Describe the scalar product of two vectors in detail, including the algebraic expression in component form, the laws it obeys, and its use in computing work done.

Derive the formula for maximum compression of a spring when a car of mass m collides with it at speed v, and explain how the result changes if friction (coefficient μ) is present on the surface.

Explain the concept of power in physics, derive the relation P = F·v, and discuss the practical significance of the units watt and horsepower with examples.

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