Question

A crate of mass m1 = 13.00 kg is pulled by a massless rope up a 36.9° ramp. The rope passes over an ideal pulley and is attached to a hanging crate of mass m2 = 16.30 kg. The crates move 1.41 m, starting from rest.Find the work done by gravity on the sliding crate.

Answer 1

The free body diagram of the system can be shown as,

The work done by gravity on sliding crate can be given as,

$W=(m_1g\sin \theta)dcos180^{\circ}_{}$

The angle between distance and force is 180 because the force acts in the downward direction whereas the block slide in the upward direction.

Substitute the known values,

$\begin{gathered} W=(13.0kg)(9.8m/s^2)\sin 36.9^{\circ}(1.41\text{ m)(-1)(}\frac{1\text{ J}}{1kgm^2s^{-2}}) \\ =-(179.634\text{ J)(}0.6) \\ \approx-107.8\text{ J} \end{gathered}$

Thus, the work done by gravity on the sliding crate is -107.8 J.