Answer
See the detailed answer below.
Work Step by Step
- The work done by the gravitational force is the easiest one to count and is given by
$$W_{F_G}=F_G\;\Delta y\;\cos\theta=F_G\Delta y\;\;\overbrace{\cos0^\circ}^{=1}$$
where $\theta$ is the angle between the force direction and the displacement direction.
Thus,
$$W_{F_G}=F_G\Delta y $$
Plugging the known;
$$W_{F_G}=( 2500)( 5)=\color{red}{\bf 12,500 }\;\rm J $$
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- The work done by the tension in the right rope;
$$W_{T_{2}}=T_2\;\Delta y\; \cos(90^\circ+45^\circ)$$
Plugging the known;
$$W_{F_G}=(1295)(5)\cos135^\circ=\color{red}{\bf -4,579}\;\rm J $$
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- The work done by the tension in the left rope;
$$W_{T_{1}}=T_1\;\Delta y\; \cos(90^\circ+60^\circ)$$
Plugging the known;
$$W_{F_G}=(1830)(5)\cos150^\circ=\color{red}{\bf -7,924}\;\rm J $$
