Work & Energy — Set 5

Correct Answer: D. 600 W

• **600 W** = P = W/t = 1200 J / 2 s = 600 W. • **P = 1200 ÷ 2 = 600 W** — the motor delivers 600 joules of energy per second; a higher power rating means faster work completion. • 💡 Wrong-option analysis: 240 W: 1200/5 = 240 — wrong divisor (5 s instead of 2 s); 1200 W: confuses W (watts) with W (work done in joules) — 1200 J is the work, not the power; 2400 W: 1200 × 2 = 2400 — multiplied instead of divided.

Correct Answer: A. -fs

• **-fs** = friction force f acts opposite to displacement s (angle = 180°); W = f × s × cos 180° = -fs. • **W_friction = -fs** — this negative work reduces the body's KE by fs; the removed energy appears as heat at the sliding surfaces. • 💡 Wrong-option analysis: +f/s: dividing f by s gives N/m — wrong dimensions for work (should be N·m = J); 0: zero work would require f ⊥ s (perpendicular), but friction is antiparallel (180°) to motion; +fs: positive would mean friction aids the motion, which contradicts the opposing nature of friction.

Correct Answer: A. Kinetic and potential energy

• **Kinetic and potential energy** = mechanical energy = KE + PE; it is the energy associated with the position and motion of a body in a mechanical system. • **E_mech = ½mv² + mgh** — mechanical energy does not include thermal, chemical, or nuclear energy; it is the macroscopic form of energy. • 💡 Wrong-option analysis: Heat and light energy: these are thermal and radiant forms — not mechanical; Mass and time: mass is not energy (unless via E = mc²) and time is not an energy form; Chemical and nuclear energy: stored in molecular or nuclear bonds — not the macroscopic kinetic/potential mechanical forms.

Correct Answer: D. K

• **K (KE ∝ v²)** = kinetic energy KE = ½mv² is proportional to the square of speed; doubling speed quadruples KE. • **KE ∝ v²** — label 'K' represents the correct relationship KE ∝ v²; this quadratic dependence is why highway accidents at high speed are so severe. • 💡 Wrong-option analysis: [Option not available]: not a valid physics option; i: represents √(-1) in mathematics — has no relevance to the KE-speed relationship; E: may suggest E ∝ v, but the actual relationship is quadratic (v²), not linear.

Correct Answer: D. 173 J

• **173 J** = W = F s cosθ = 50 × 4 × cos 30° = 50 × 4 × 0.866 = 173.2 J ≈ 173 J. • **W = 50 × 4 × 0.866 ≈ 173 J** — cos 30° = √3/2 ≈ 0.866; only this component of the 50 N force acts along the 4 m displacement. • 💡 Wrong-option analysis: 0 J: zero requires cos θ = 0, i.e. θ = 90°; here θ = 30° so there is a significant work component; 100 J: 50 × 4 × 0.5 = 100 — uses cos 60° = 0.5 instead of cos 30° = 0.866; 200 J: 50 × 4 × 1 = 200 — ignores the angle entirely (uses cos 0° = 1).

Correct Answer: D. Kinetic energy

• **Kinetic energy** = during free fall, height decreases so mgh decreases; by energy conservation, the lost PE converts to KE (ignoring air resistance). • **ΔPE = -ΔKE** — the decrease in gravitational PE exactly equals the increase in KE; at the moment of hitting the ground, all PE has become KE. • 💡 Wrong-option analysis: Nuclear energy: nuclear energy involves changes in atomic nuclei — free fall does not trigger nuclear reactions; Chemical energy: chemical bonds are not involved in gravitational free fall; Sound energy: some sound is produced on impact, but during the fall itself (ignoring air), no sound energy is generated.

Correct Answer: A. Increase in potential energy

• **Increase in potential energy** = when lifting slowly, acceleration ≈ 0 so ΔKE ≈ 0; by the work-energy theorem, all work against gravity becomes gravitational PE. • **W_lift = ΔPE = mgh** — lifting at constant velocity (KE unchanged) means every joule of work input equals one joule of PE gained. • 💡 Wrong-option analysis: Change in time: time is not an energy quantity — work does not equal Δt; Decrease in kinetic energy: KE decreases when negative work is done; slow lifting does positive work that stores PE, not reduces KE; Increase in momentum: impulse-momentum theorem involves force × time, not work × displacement.

Correct Answer: B. P = Fv

• **P = Fv** = when force is parallel to velocity, instantaneous power = force × speed; since θ = 0°, P = F v cos 0° = Fv. • **P = F · v** — this formula allows calculation of engine power at any instant; e.g., a car engine exerting 2000 N at 30 m/s delivers 60 000 W = 60 kW. • 💡 Wrong-option analysis: P = F/v: dividing gives N/(m/s) = N·s/m — not watts (N·m/s = W); P = F + v: addition of force (N) and speed (m/s) is dimensionally invalid; P = v/F: gives m/(s·N) = m·s/kg — dimensionally wrong for power.

Correct Answer: D. 2000 W

• **2000 W** = P = Fv = 1000 N × 2 m/s = 2000 W; the pump must exert 1000 N continuously while moving water at 2 m/s. • **P = 1000 × 2 = 2000 W** — this is the minimum power needed to sustain the water flow; additional power would be needed in practice to overcome pipe friction. • 💡 Wrong-option analysis: 3000 W: 1000 × 3 = 3000 — uses v = 3 m/s, not 2 m/s; 500 W: 1000/2 = 500 — division instead of multiplication; 1000 W: this is just the force value — speed (2 m/s) was not multiplied.

Correct Answer: A. 1 J/s

• **1 J/s** = watt is the SI unit of power; 1 W = 1 joule per second, meaning 1 joule of energy is transferred or 1 joule of work is done per second. • **1 W = 1 J/s = 1 kg·m²/s³** — this unit definition makes it easy to convert between power (watts) and energy (joules) once time is known. • 💡 Wrong-option analysis: 1 J·s: joule-second is the unit of action (Planck's constant has this unit) — not power; 1 N·m: newton-metre is a joule (unit of energy/work), not power (rate of work); 1 kg·m²/s²: this is one joule — the unit of energy, not power.