Work & Energy — Set 6

Correct Answer: C. A 1 N force moves a body by 1 m in its direction

• **A 1 N force moves a body by 1 m in its direction** = definition of 1 joule: W = 1 N × 1 m × cos 0° = 1 J. • **1 J = 1 N·m** — this links the unit of work to the units of force and distance; it is the fundamental definition from which all energy units derive. • 💡 Wrong-option analysis: A 1 N force acts for 1 s: force × time = impulse (N·s), not work; A 1 kg mass is lifted by 1 m: this is mgh = 1 × g × 1 = g joules ≈ 9.8 J — not 1 J; A 1 W power acts for 1 hour: 1 W × 3600 s = 3600 J = 1 Wh — not 1 J.

Correct Answer: D. Area under the force–displacement graph

• **Area under the force–displacement graph** = for a variable force, W = ∫F dx; graphically this equals the area between the F–x curve and the x-axis. • **W = ∫F dx** — for a spring (F = kx), the F–x graph is a triangle; its area = ½ × base × height = ½ × x × kx = ½kx², confirming the spring energy formula. • 💡 Wrong-option analysis: Slope of the force–displacement graph: dF/dx = k (spring constant) — this gives the stiffness, not the work; Area under the velocity–time graph: that area gives displacement (∫v dt = Δx), not work; Slope of the displacement–time graph: that slope gives velocity (dx/dt = v), not work.

Correct Answer: D. 10 m/s

• **10 m/s** = to convert km/h to m/s, divide by 3.6: 36 ÷ 3.6 = 10 m/s. • **÷ 3.6** — equivalently multiply by 1000/3600 = 5/18; 36 × 5/18 = 180/18 = 10 m/s. • 💡 Wrong-option analysis: 12 m/s: 43.2 km/h ÷ 3.6 = 12 — this would be for 43.2 km/h, not 36 km/h; 36 m/s: confuses km/h with m/s — this ignores the unit conversion entirely; 6 m/s: 21.6 km/h ÷ 3.6 = 6 — this would be for 21.6 km/h, not 36 km/h.

Correct Answer: B. 20 J

• **20 J** = KE = ½mv² = ½ × 10 × (2)² = ½ × 10 × 4 = 20 J. • **KE = 0.5 × 10 × 4 = 20 J** — note 2² = 4 and ½ × 10 = 5; 5 × 4 = 20 J. • 💡 Wrong-option analysis: 40 J: this is mv² = 10 × 4 = 40 — missing the ½ factor; 10 J: ½ × 10 × 2 = 10 — uses v = 2 instead of v² = 4; 5 J: ½ × 10 × 1 = 5 — uses v² = 1, treating v = 1.

Correct Answer: C. PE = mgh

• **PE = mgh** = near Earth's surface where g is approximately constant, gravitational potential energy is the product of mass, gravitational acceleration, and height. • **PE = mgh** — m in kg, g = 9.8 or 10 m/s², h in metres; PE is linear in both m and h; this approximation is valid up to heights much smaller than Earth's radius. • 💡 Wrong-option analysis: PE = p²/(2m): this is kinetic energy expressed in terms of momentum — the formula for KE, not PE; PE = (1/2)mv²: this is the kinetic energy formula — not gravitational PE; PE = F/t: F/t has dimensions of power rate, not energy.

Correct Answer: A. 200 J

• **200 J** = W = mgh = 2 × 10 × 10 = 200 J (using g = 10 m/s²). • **W = 2 × 10 × 10 = 200 J** — this work equals the gain in gravitational PE stored in the mass when raised to 10 m. • 💡 Wrong-option analysis: 1000 J: this could be 10 × 10 × 10 = 1000 — using the wrong mass of 10 kg instead of 2 kg; 20 J: 2 × 10 = 20 — the height (10 m) was not multiplied in; 100 J: 2 × 10 × 5 = 100 — uses h = 5 m instead of 10 m, or 1 × 10 × 10 using wrong mass.

Correct Answer: A. Maximum kinetic energy

• **Maximum kinetic energy** = at the lowest (mean) position of a pendulum, height is minimum so PE is minimum; by energy conservation, KE is at its maximum. • **KE_max at bottom** — total mechanical energy = KE + PE = constant; when PE is minimum (height = 0 from reference), KE must be maximum. • 💡 Wrong-option analysis: Maximum potential energy: PE is maximum at the extreme (highest) positions where the bob momentarily stops; Zero total energy: total mechanical energy is conserved and non-zero — it equals the PE at the highest point; Zero kinetic energy: KE is zero at the extreme positions, not at the lowest point.

Correct Answer: B. 100%

• **100%** = efficiency = (useful output / input) × 100%; since useful output ≤ input (energy cannot be created), efficiency can never exceed 100%. • **η ≤ 100%** — real machines always have losses (friction, heat, sound) so actual efficiency < 100%; 100% is the theoretical upper limit. • 💡 Wrong-option analysis: 90%: 90% is a good practical efficiency but it is not the universal maximum — some machines exceed 90%; 50%: half is not the maximum — many machines operate above 50% efficiency; 75%: 75% is an arbitrary value; the absolute limit set by energy conservation is 100%.

Correct Answer: C. Square of extension

• **Square of extension** = elastic PE = ½kx²; PE is proportional to x², so if x doubles, PE quadruples. • **PE ∝ x²** — the quadratic relationship means small extensions store little energy but large extensions store much more; this has implications for spring safety. • 💡 Wrong-option analysis: Extension: PE ∝ x would mean spring energy increases linearly — but doubling x only doubles PE, contradicting the observed quadrupling; Inverse of extension: PE ∝ 1/x would mean compressing the spring more stores less energy — physically wrong; Cube of extension: PE ∝ x³ has no basis in Hooke's Law.

Correct Answer: D. 1 kWh

• **1 kWh** = 100 W = 0.1 kW; energy = 0.1 kW × 10 h = 1 kWh. • **E = 0.1 kW × 10 h = 1 kWh** — in SI units, 1 kWh = 3.6 × 10⁶ J; this is the unit used by electric utility companies for billing. • 💡 Wrong-option analysis: 10 kWh: this uses 1 kW × 10 h — fails to convert 100 W to 0.1 kW; 0.1 kWh: this uses 0.1 kW × 1 h — uses only 1 hour instead of 10 hours; 100 kWh: 100 W × 10 h = 1000 Wh — in kWh this is 1, not 100.