Calculate work done, kinetic and potential energy, and the rate energy is transferred or converted — nearly everything here connects back to one idea, the conservation of energy, tracked as it changes form. Every question comes with a written walkthrough of exactly where the energy went.
W = Fd only applies directly when the force is in the same direction as the motion — a force applied at an angle only contributes its component along the direction of motion, which is why the full formula includes cos(θ).
Conservation of mechanical energy (kinetic plus potential) only holds when no external forces like friction remove energy from the system — with friction present, some mechanical energy converts to heat and the total mechanical energy decreases.
Work measures the total energy transferred, while power measures how quickly that transfer happens — two situations can involve the same amount of work but very different power if one takes much longer than the other.
Straight from the bank — one per difficulty tier. Reveal the answer to see the explanation you'd get in a real session.
A 20 kg box is lifted 3 meters straight up at constant velocity. Using g = 10 m/s², how much work is done against gravity?
A — 600 J. The force needed to lift the box at constant velocity equals its weight: F = mg = (20)(10) = 200 N. Work done: W = Fd = (200)(3) = 600 J. 60 J (B) comes from forgetting to include gravity — computing (20)(3) as if mass alone were the force, without multiplying by g.
A 4 kg object is moving at 6 m/s. What is its kinetic energy?
A — 72 J. Kinetic energy: KE = ½mv² = ½(4)(6²) = ½(4)(36) = 72 J. 144 J (B) comes from forgetting the ½ factor — computing mv² directly as 4×36=144 without halving it.
A 2 kg ball is dropped from a height of 10 meters. Using g = 10 m/s², and ignoring air resistance, what is the ball's speed just before it hits the ground?
A — ≈14.14 m/s. By conservation of energy, all the gravitational potential energy converts to kinetic energy by the time the ball hits the ground: mgh = ½mv². The mass cancels out, leaving v² = 2gh = 2(10)(10) = 200, so v = √200 ≈ 14.14 m/s. 20 m/s (B) comes from computing v² as 4gh instead of 2gh — doubling the correct value under the square root, a plausible-looking variant of the formula.
Work, energy, and power problems reward tracking where energy goes — from potential to kinetic, or from work done to power delivered over time — rather than memorizing formulas in isolation.
For every problem, state which form of energy is present at the start and what it converts into by the end. Untimed practice is where that tracking habit sticks.
Move to timed sessions once the work, kinetic energy, and potential energy formulas are instant recall.
Pair work, energy, and power with kinematics in a mock — both describe the same motion from different angles.
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