There is a ball of mass 320 g. It has 625 J potential energy when released freely from a height. The speed with which it will hit the ground is
- (a)62.5 m/s
- (b)20 m/s
- (c)50 m/s
- (d)40 m/s
Correct — A, 62.5 m/s. In free fall (ignoring air resistance) all the gravitational potential energy converts to kinetic energy, so at the ground KE = PE = 625 J. With m = 320 g = 0.32 kg, (1/2) x m x v^2 = 625 gives v^2 = 2 x 625 / 0.32 = 3906.25, so v = sqrt(3906.25) = 62.5 m/s.
- (b)20 m/s — Far too small: (1/2) x 0.32 x 20^2 = 64 J, nowhere near the given 625 J.
- (c)50 m/s — Gives (1/2) x 0.32 x 50^2 = 400 J, still short of 625 J.
- (d)40 m/s — Gives (1/2) x 0.32 x 40^2 = 256 J, far below 625 J.
Conservation of mechanical energy in free fall: the potential energy a body has at a height is fully converted into kinetic energy by the time it reaches the ground (air resistance neglected). So KE at the ground = PE at the top = (1/2) m v^2. Solve for v after converting mass to kilograms.
The trap is unit handling: use the mass as 320 (kg) instead of 0.32 kg and the speed comes out badly wrong; the height and g never need to be known because energy conservation ties PE directly to KE.
- Conservation of energy in free fall: loss in PE = gain in KE, i.e. mgh = (1/2) m v^2.
- KE = (1/2) x m x v^2; with KE = 625 J and m = 0.32 kg, v^2 = 2 x 625 / 0.32 = 3906.25.
- sqrt(3906.25) = 62.5, so v = 62.5 m/s.
- 320 g = 0.32 kg (divide grams by 1000).
- Using the mass in grams instead of kilograms.
- Dropping the factor of one-half in KE = (1/2) m v^2.
Asked as an energy-conservation numeric: potential energy at a height equals kinetic energy at the ground; solve for the speed.
The planet Mercury is revolving in an elliptical orbit around the sun as shown in the given figure. The kinetic energy of Mercury is greatest at the point labelled
- (a) A
- (b) B
- (c) C
- (d) D
Answer(a) A
Same core idea of kinetic energy and energy conservation: KE is greatest where speed is greatest (perihelion, A) as energy is exchanged in orbit, just as this NDA item has potential energy convert fully into kinetic energy during a free fall.
- practice — not a real PYQ
A body of mass 2 kg has 100 J of kinetic energy. Its speed is
- (a)10 m/s
- (b)50 m/s
- (c)100 m/s
- (d)5 m/s
Answer(a) 10 m/s — (1/2) x 2 x v^2 = 100 gives v^2 = 100, v = 10.
- practice — not a real PYQ
A body falls freely from a height h. At the instant it reaches the ground, its kinetic energy is equal to
- (a)half its initial potential energy
- (b)its initial potential energy
- (c)twice its initial potential energy
- (d)zero
Answer(b) its initial potential energy — energy is conserved, so all PE becomes KE.