A force of F = (5y + 20)ĵ N acts on a particle. The work done by this force when the particle is moved from y = 0 m to y = 10 m is ___________ J.

Solution

Physics sir · 4 steps

The force changes with y, so do not multiply force by distance. Integrate it over the path.

1. Set up the integral. The force acts along ĵ and the particle moves along y, so dW = F dy. The force depends on y, so add up these small pieces of work from y = 0 to y = 10 m.

   W = ∫₀¹⁰ (5y + 20) dy = [5y²/2 + 20y]₀¹⁰

2. Put in limits. At y = 10: (5 × 100)/2 + 20 × 10 = 250 + 200 = 450. At y = 0 the bracket is 0.
3. Quick check. The force rises in a straight line from 20 N to 70 N, so the area under the F–y graph is a trapezium: (20 + 70)/2 × 10 = 450.
4. So. The work done is 450 J.

The common slip: 700. Treating the force as constant at its final value, 70 × 10, or forgetting the 1/2 when integrating 5y, gives 700, not 450.

The idea to keep. When the force depends on position, work is the area under the F versus displacement graph, so use W = ∫ F dx. If F is linear, the average of the end values times the displacement gives the same answer.

Answer from the NTA final answer key. Solution by PrepWiser.