Q1. A 2 kg ball falls from a height of 10 m. Find its speed just before hitting the ground (g = 10).
PE at top = mgh = 2×10×10 = 200 J → all becomes KE
½mv² = 200 ⇒ v² = 400/2 × ... ⇒ ½×2×v² = 200
v² = 200 ⇒ v ≈ 14.1 m/s
Work transfers energy; power is how fast. Compute work, energy and power for any situation and practise the standard numericals.
Change force, mass, speed and time to see work, kinetic energy and power update (g = 10 m/s²).
Kinetic energy grows with the SQUARE of speed — doubling speed quadruples the energy (and stopping distance).
Work, energy and power problems with steps (g = 10 m/s²).
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Q1. A 2 kg ball falls from a height of 10 m. Find its speed just before hitting the ground (g = 10).
PE at top = mgh = 2×10×10 = 200 J → all becomes KE
½mv² = 200 ⇒ v² = 400/2 × ... ⇒ ½×2×v² = 200
v² = 200 ⇒ v ≈ 14.1 m/s
Q2. A pump raises 100 kg of water to a height of 10 m in 20 s. Find its power (g = 10).
Work = mgh = 100 × 10 × 10 = 10000 J
P = W/t = 10000 / 20
P = 500 W
| Quantity | Formula | SI unit |
|---|---|---|
| Work | W = F s cosθ | joule (J) |
| Kinetic energy | KE = ½ m v² | J |
| Potential energy | PE = m g h | J |
| Power | P = W / t = F v | watt (W) |
| Work–energy theorem | W_net = ΔKE | J |
The net work done on a body equals the change in its kinetic energy: W = ΔKE.
When there is no displacement, or when the force is perpendicular to the displacement (θ = 90°).
Energy (joule) is the capacity to do work; power (watt) is the rate at which work is done, P = W/t.
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