Question

A block of mass 5.0 kg slides down an incline of inclination 30° and length 10 m. Find the work done by the force of gravity.

Answer 1

$$\begin{gathered} \mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},M=5\mathrm{kg} \\ \mathrm{Angle}\mathrm{of}\mathrm{inclination},\theta =30° \end{gathered}$$
Gravitational force acting on the block, $\mathrm{F}=mg$
Work done by the force of gravity depends only on the height of the object, not on the path length covered by the object.


$$\begin{gathered} \mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{object},h=10\times \sin 30° \\ =10\times \frac{1}{2}=5\mathrm{m} \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Work}\mathrm{done}\mathrm{by}\mathrm{the}\mathrm{force}\mathrm{of}\mathrm{gravity},w=mgh \\ =5\times 9.8\times 5=245\mathrm{J} \end{gathered}$$