Law of Conservation of Energy MCQs

Q1. A ripe mango of mass 3 kg drops off a branch from 6 m above the ground. Ignoring air resistance, with what speed does it reach the ground?

• A.7.746 m/s
• B.120 m/s
• C.10.954 m/s
• D.60 m/s

Answer: C. 10.954 m/s

Conservation of energy: (1/2)m*v^2 = m*g*h, so v = sqrt(2*g*h) = sqrt(2*10*6) = sqrt(120) = 10.954 m/s. The mass cancels out.

Q2. A bricklayer's hammer of mass 0.2 kg slips from a scaffold from 8 m up. What is its kinetic energy when it has fallen to a point 6 m above the ground?

• A.16 J
• B.0.4 J
• C.4 J
• D.12 J

Answer: C. 4 J

It has fallen (h - hf) = 8 - 6 = 2 m. The PE lost over that drop equals the KE gained: KE = m*g*(h-hf) = 0.2*10*2 = 4 J.

Q3. A flowerpot of mass 0.5 kg topples from a balcony railing from a height of 18 m. Using conservation of energy, what is its kinetic energy the instant before it lands?

• A.4.5 J
• B.5 J
• C.9 J
• D.90 J

Answer: D. 90 J

At the top all energy is potential: PE = m*g*h = 0.5*10*18 = 90 J. By conservation of energy this entire amount becomes kinetic energy at the bottom, so KE = 90 J.