A $15\ \mathrm{kg}$ crate is moved along a horizontal floor by a warehouse worker who’s pulling on it with a rope that makes a $30^\circ $ angle with the horizontal. The tension in the rope is $69\ \mathrm{N}$, and the crate slides a distance of $10\ \mathrm{m}$.
How much work is done on the crate by the worker?
$W=600\ \mathrm{J}$
The figure below shows that $\textbf F_{\mathrm{T}}$ and $\textbf d$ are not parallel. It’s only the component of the force acting along the direction of motion, $F_{\mathrm{T}}\cos \theta $, that does work.
Therefore,
$$W=(F_{\mathrm{T}}\cos{\theta })d$$
$$W=(69\ \mathrm{N\cdot \cos{30^\circ })(10\ m)}$$
$$W=600\ \mathrm{J}$$