Work & Energy — Set 2

Correct Answer: B. [M L^2 T^-2]

• **[M L² T⁻²]** = energy = work = force × distance; dimensions are [M L T⁻²] × [L] = [M L² T⁻²]. • **Dimensional analysis** — joule = N·m = (kg·m/s²)·m = kg·m²/s²; the SI units confirm [M L² T⁻²]. • 💡 Wrong-option analysis: [M L² T⁻³]: this is the dimensional formula for power (W = J/s), not energy; [M L T⁻²]: this is the formula for force (Newton), not energy; [M L T⁻¹]: this is the formula for momentum (kg·m/s), not energy.

Correct Answer: D. (Useful output energy / Input energy) x 100%

• **(Useful output energy / Input energy) × 100%** = efficiency compares how much of the input energy is converted to useful work; always ≤ 100% due to losses. • **η = (W_out / W_in) × 100%** — losses like heat from friction reduce the useful output; a 100% efficient machine is an ideal concept only. • 💡 Wrong-option analysis: (Input/Useful output) × 100%: inverting numerator and denominator gives a number ≥ 100%, which is not efficiency; Input energy - Output energy: this is the energy lost to waste, not efficiency; (Output power × Input power) × 100%: multiplying the two power values is dimensionally wrong for defining efficiency.

Correct Answer: A. 100 J

• **100 J** = work done lifting a mass against gravity: W = mgh = 5 × 10 × 2 = 100 J. • **W = 5 × 10 × 2 = 100 J** — all this work is stored as gravitational potential energy in the mass-Earth system. • 💡 Wrong-option analysis: 10 J: this is m × g = 5 × 10 / 5 — the height (2 m) was not included; 20 J: this is m × h = 5 × 2 × 2 or just m × h not including g; 50 J: this is (m × g)/1 = 5 × 10 ÷ 1 — the height (2 m) was divided instead of multiplied.

Correct Answer: C. Negative

• **Negative** = friction acts opposite to displacement (θ = 180°), so W = f × s × cos 180° = -fs; friction removes kinetic energy from the body. • **W_friction = -fs** — negative work means the body's kinetic energy decreases by fs; this energy is converted to heat at the contact surfaces. • 💡 Wrong-option analysis: Always zero: zero work requires perpendicular force or zero displacement; friction does remove energy; Always positive: positive work would mean friction helps the motion, which contradicts its opposition to motion; Always maximum: there is no maximum constraint on friction's (negative) work.

Correct Answer: D. 746 W

• **746 W** = one horsepower equals approximately 746 watts; originally defined as the rate a horse could work over a sustained period. • **1 hp = 745.7 W ≈ 746 W** — this conversion is essential for interpreting engine and motor specifications in both SI and imperial units. • 💡 Wrong-option analysis: 100 W: 100 W is a common power for a light bulb — far less than one horsepower; 76 W: this is often mistakenly quoted as metric horsepower (1 PS = 735.5 W); not the standard definition; 7460 W: this would be 10 horsepower — ten times the correct value.

Correct Answer: A. Kinetic energy

• **Kinetic energy** = in a perfectly elastic collision, both momentum and kinetic energy are conserved; no energy is lost to deformation or heat. • **KE_before = KE_after** — for elastic collision: ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂²; billiard ball collisions approximate this. • 💡 Wrong-option analysis: Mechanical power: power is a rate (J/s) and is not a conserved quantity in collisions; Temperature: temperature may change locally at the impact point but is not conserved in collisions; Time: time is not a conserved quantity — it is the independent variable.

Correct Answer: B. 2 kWh

• **2 kWh** = electrical energy = power in kW × time in hours = 1 kW × 2 h = 2 kWh. • **Energy = P × t = 1 × 2 = 2 kWh** — 1 kWh = 3.6 × 10⁶ J; kWh is the unit on your electricity bill. • 💡 Wrong-option analysis: 1 kWh: this would require the device to run for only 1 hour at 1 kW, not 2 hours; 0.5 kWh: this is 1 kW × 0.5 h — half the correct time; 2000 kWh: this mixes units — perhaps 1 kW × 2000 h or uses 1 W × 2000 h, neither correct for 2 hours.

Correct Answer: A. Zero

• **Zero** = centripetal force always points toward the centre while displacement is tangential; force is perpendicular to displacement, so W = Fv cos 90° = 0. • **θ = 90° always** — at every instant in uniform circular motion, force ⊥ velocity, confirming zero instantaneous power and hence zero work over any arc. • 💡 Wrong-option analysis: Positive: positive work would increase KE, but speed is constant in uniform circular motion — consistent with zero work; Negative: negative work would decrease KE and slow the object, but speed stays constant; Maximum: work has no defined maximum here — it is exactly zero at every instant.

Correct Answer: D. A chosen reference level

• **A chosen reference level** = gravitational PE = mgh is defined relative to an arbitrarily chosen reference level where PE = 0; only differences in PE are physically meaningful. • **Arbitrary zero** — the ground, the table surface, or the lowest point can be the reference; choosing different levels changes PE values but not ΔPE, which alone determines physics. • 💡 Wrong-option analysis: The lowest point always: for a problem set on a mountain, the base of the mountain might be chosen — not necessarily Earth's lowest point; The center of Earth always: PE = 0 at Earth's centre would make calculations on the surface very inconvenient; The highest point always: making the top the zero level is allowed but not standard — the reference is arbitrary.

Correct Answer: C. W = F s cosθ

• **W = F s cosθ** = work done by a constant force F causing displacement s at angle θ between force and displacement directions. • **Scalar dot product** — W = F⃗ · s⃗ = Fs cosθ; only the component of force along displacement (F cosθ) contributes; the perpendicular component does no work. • 💡 Wrong-option analysis: W = mgh/t: mgh/t is power (for slow vertical lift), not work; W = F/t: F/t has dimensions of impulse rate — not work; W = (1/2)mv: this has wrong dimensions (momentum is mv, KE is ½mv²) — neither equals work generally.