A 10 kg crate is moved across a horizontal floor by a worker who applies a constant horizontal force of 50 N. The crate travels 5 m. If the coefficient of kinetic friction between crate and floor is 0.2, find the work done by the worker on the crate.

Step-by-step solution

1. Compute the kinetic friction force: $f_k={\mu }_{k}mg=0.2\times 10\times 9.8=19.6\text{N}$.
2. The force exerted by the worker is applied in the direction of displacement, so the angle $\theta =0^{\circ }$ and $cos\theta =1$.
3. Calculate the work done by the worker: $W_{\text{worker}}=Fdcos\theta =50\times 5\times 1=250\text{J}$.
4. The result is the work done by the worker; friction does negative work but is not subtracted because the question asks only for the worker’s work.

Concept tested: Work done by a constant force

Published 4 October 2026 · How this solution was prepared and checked