A 15‑kg cart is pulled uphill by a horizontal force that increases linearly from 10 N at the bottom to 20 N at the top over a distance of 12 m. The hill makes a 30° angle with the horizontal. What is the work done by the pulling force?

Step-by-step solution

1. Express the force as a function of distance: $F(x)=10+\frac{10}{12}x=10+\frac{5}{6}x$ N.
2. Integrate to find the horizontal work: $W_{\text{hor}}={\int }_0^{12}F(x),dx=180$ J.
3. Since the displacement is along the hill, the actual displacement is $Deltas=12$ m along the slope and the horizontal component is $Deltascos30^{c}irc$.
4. Multiply the horizontal work by $cos30^{\circ }$ to account for the slope: $W=W_{\text{hor}}cos30^{\circ }=180\times \frac{\sqrt{3}}{2}\approx 155.9$ J.
5. Rounded to the nearest integer, the work done by the pulling force is 156 J.

Concept tested: Work done by a variable force on an incline

Published 7 October 2026 · How this solution was prepared and checked