Net Work Equals Change in Kinetic Energy: Pick the Easiest Route to It
Work questions come in three disguises: a dot product of force and displacement, an area or integral for a changing force, and the work-energy theorem, which says the net work equals the change in kinetic energy. The theorem is usually the shortest path.
Work Done and the Work-Energy Theorem is the central idea in this question pattern. Identify the condition in the stem before applying the method below.
Work as a dot product, an integral or an area
W = F · s uses only the component of force along the displacement, and the displacement is the final position minus the initial one: for F = (2, 3, 0) N and a move from (-4, 6, 0) to (0, 5, 3), s = (4, -1, 3) and W = 8 - 3 = 5 J.
On an F-x graph, W is the area under the curve, and areas below the axis count negative. A force that pushes down at an angle on a sliding block raises the normal force to mg + F sinθ, so the work done by friction over a distance S is -μ(mg + F sinθ) S.
The work-energy theorem
For x = 3t - 4t² + t³ the velocity v = 3 - 8t + 3t² goes from 3 m/s at t = 0 to 19 m/s at t = 4 s, so a 30 g particle receives (1/2)(0.03)(19² - 3²) = 5.28 J.
Dissipated energy and variable mass
For a ball thrown up at 14 m/s that reaches only 8.0 m, the loss to air drag is (1/2)(0.5)(14)² - (0.5)(9.8)(8) = 49 - 39.2 = 9.8 J. For a 1 g raindrop falling 1 km and landing at 50 m/s, gravity does 10 J and the net gain in kinetic energy is (1/2)(0.001)(50)² = 1.25 J, so air resistance does -8.75 J. For a leaking bucket pulled up at constant speed, the mass falls steadily from 15 kg to 9 kg, so use the average mass 12 kg: W = 12 x 10 x 15 = 1800 J. Starting from rest, a body accelerated uniformly to speed v in time t1 has done W(t) = (1/2) m (v/t1)² t² by time t.
OlympIQ exam lens
- ✓W = F · s with s = final - initial position; W = ∫F dx = area under the F-x graph
- ✓Net work = ΔK, which gives work done by all forces together
- ✓Work by gravity = m g x (vertical drop); the path is irrelevant
- ✓Energy lost to drag = ΔK + ΔU mismatch, i.e. (1/2)mv² - mgh for an ascent
- ✓A varying mass loses at a constant rate: use the average mass
Do not multiply the force by the path length when the force is not along the displacement. Only the component along the displacement does work, and the displacement is a vector from the initial to the final position, not the length of the path travelled.
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