\(F - Mg sin 30° = 0 \rightarrow F = Mg sin 30°\) અને \(d = 10 m\)
\({\text{W = Fdcos}}\theta {\text{ = (Mgsin3}}{{\text{0}}^ \circ }{\text{ ) dcos}}{{\text{0}}^ \circ }{\text{ = 10 }} \times {\text{ 10 }} \times \frac{{\text{1}}}{{\text{2}}}\, \times \,{\text{ 10 }} \times {\text{ 1 = 500 J}}\)