Work & Energy — Set 1

Correct Answer: C. Joule

• **Joule (J)** = the SI unit of work and energy; defined as the work done when a force of 1 newton displaces an object by 1 metre in the direction of the force. • **1 J = 1 N m = 1 kg m²/s²** — this composite unit shows how work links force (kg m/s²) and distance (m). • 💡 Wrong-option analysis: Newton: the unit of force, not energy or work; Pascal: the unit of pressure (N/m²), not energy; Watt: the unit of power (J/s), which is rate of doing work, not the work itself.

Correct Answer: A. 50 J

• **50 J** = work is calculated using W = F s cosθ; here F = 10 N, s = 5 m, and θ = 0° (force and displacement in the same direction). • **W = 10 × 5 × cos 0° = 50 J** — cos 0° = 1, so all the force contributes to work; no component is wasted. • 💡 Wrong-option analysis: 2 J: this would be F/s = 10/5 — division, not multiplication, which is wrong; 5 J: this is just the displacement value; the formula requires F × s, not s alone; 15 J: this is F + s = 10 + 5 — addition has no place in the work formula.

Correct Answer: D. Four times

• **Four times** = kinetic energy KE = ½mv²; since KE is proportional to v², doubling v multiplies v² by 4, so KE becomes 4 times. • **KE ∝ v²** — if v → 2v, then KE → ½m(2v)² = ½m × 4v² = 4 × ½mv²; the quadratic dependence is key. • 💡 Wrong-option analysis: Double: this would be true only if KE ∝ v (linear), but KE ∝ v²; Half: halving would occur if speed were halved, not doubled; Triple: there is no physical scenario that triples KE when speed doubles — v² scaling gives exactly 4×.

Correct Answer: C. When force is perpendicular to displacement

• **When force is perpendicular to displacement** = W = F s cosθ; at θ = 90°, cos 90° = 0, making work done zero regardless of the magnitudes of F and s. • **W = F s cos 90° = 0** — the centripetal force in uniform circular motion is the classic example: force points inward but displacement is tangential. • 💡 Wrong-option analysis: When speed is maximum: at maximum speed the body is still moving and a net force can still do work; when force is maximum: a large force can still do zero work if it is perpendicular to displacement; when energy is maximum: total energy being maximum is unrelated to whether the instantaneous work done is zero.

Correct Answer: B. 200 W

• **200 W** = power P = W/t = 600 J / 3 s = 200 W; power is the rate at which work is done or energy is transferred. • **P = W/t** — 600 ÷ 3 = 200; a higher power means the same work is completed in less time. • 💡 Wrong-option analysis: 150 W: 600/4 = 150 — wrong divisor; 600 W: confusing P with W — 600 J is the work done, not the power; 1800 W: 600 × 3 = 1800 — this is W × t, not W/t.

Correct Answer: C. Double

• **Double** = gravitational potential energy PE = mgh; when h is doubled (m and g constant), PE doubles proportionally. • **PE ∝ h** — linear dependence on height means doubling h exactly doubles PE; compare with KE which depends on v². • 💡 Wrong-option analysis: Zero: PE = mgh is only zero when h = 0 (at the reference level) — doubling h cannot make it zero; Half: halving PE would require h to be halved, not doubled; Four times: PE ∝ h (linear), not h²; four-fold increase comes from doubling a squared quantity like spring extension.

Correct Answer: A. 0°

• **0°** = work W = F s cosθ is maximum when cosθ is maximum; the maximum value of cosθ is 1, achieved at θ = 0°. • **cos 0° = 1** — when force is exactly parallel to displacement, the full magnitude of the force contributes to work; no component is wasted. • 💡 Wrong-option analysis: 180°: cos 180° = -1, giving minimum (most negative) work — pulling backwards while object moves forward; 60°: cos 60° = 0.5, giving only half the maximum possible work; 90°: cos 90° = 0, giving zero work — force is entirely perpendicular to motion.

Correct Answer: A. Depends only on initial and final positions

• **Depends only on initial and final positions** = for a conservative force (like gravity or spring force), the work done is path-independent and equals the negative change in potential energy. • **W = -(ΔPE)** — this path-independence allows definition of a potential energy function; gravity is conservative, friction is not. • 💡 Wrong-option analysis: Depends on speed only: speed is kinematic; conservative nature is about path-independence of work; Depends on time taken: conservative forces are time-independent — the same displacement from A to B always gives the same work; Is always negative: gravity does positive work on a falling body — not always negative.

Correct Answer: D. 9 J

• **9 J** = KE = ½mv² = ½ × 2 × (3)² = ½ × 2 × 9 = 9 J. • **KE = ½ × 2 × 9 = 9 J** — note the correct squaring of velocity: 3² = 9, then multiply by ½ × 2 = 1. • 💡 Wrong-option analysis: 18 J: this is m × v² = 2 × 9 — missing the ½ factor; 6 J: this is m × v = 2 × 3 — using v instead of v² and omitting ½; 12 J: this could come from ½ × 2 × 12 — incorrectly computing 3² as 12 instead of 9.

Correct Answer: D. Change in its kinetic energy

• **Change in its kinetic energy** = the work-energy theorem: W_net = ΔKE = KE_final - KE_initial; all net work goes into changing kinetic energy. • **W_net = ½mv_f² - ½mv_i²** — this result follows directly from Newton's second law integrated over displacement; it bypasses the need to find time. • 💡 Wrong-option analysis: Its potential energy: net work equals ΔKE, not ΔPE; potential energy changes are due to conservative forces; Its total energy: total energy includes PE; W_net = ΔKE, not Δ(total energy); Change in its momentum: impulse-momentum theorem (F×t = Δp) relates force and time to momentum change, not work and displacement.