A force F=(5+3 y²) acts on a particle in the y-direction, where F is in newton and y is in meter. The work done by the force during a displacement from y=2 m to y=5 m is ___________ J.

Solution

Physics sir · 3 steps

The force changes with y, so you cannot multiply force by distance. Integrate F dy from 2 to 5.

1. Variable force. Here F depends on y, so it has a different value at every point of the path. The work is the sum of F dy over the whole path, and that sum is an integral.

   W = ∫₂⁵ (5 + 3y²) dy = [ 5y + y³ ]₂⁵

2. Put in limits. At y = 5: 5(5) + 5³ = 25 + 125 = 150. At y = 2: 5(2) + 2³ = 10 + 8 = 18.
3. So. W = 150 - 18 = 132 J.

The common slip: 150. Putting in only the upper limit and forgetting to subtract the value at y = 2 gives 150, not 132.

The idea to keep. Use W = F × d only when the force is constant. When F depends on position, integrate it: W = ∫ F dy between the two positions.

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