Question

A block of weight 100 N is slowly moved up a smooth incline of inclination 37° by a person. Calculate the work done by the person in moving the block through a distance of 2 m, if the driving force is (a) parallel to the incline and (b) in the horizontal direction.

Answer 1

Given:
$$\begin{gathered} Weight,mg=100\mathrm{N} \\ \mathrm{\theta }=37°\mathrm{and}s=2\mathrm{m} \\ \mathrm{Force},\mathrm{F}=mg\sin 37° \\ =100\times \frac{3}{5}=60\mathrm{N} \end{gathered}$$
So, work done when the force is parallel to incline,
$$\begin{gathered} W=FS\cos \mathrm{\theta } \\ =60\times 2\times \cos 0°=120\mathrm{J} \end{gathered}$$




$$\begin{gathered} \mathrm{In}\Delta \mathrm{ABC},\mathrm{AB}=2\mathrm{m} \\ \mathrm{AC}=h \\ =s\times \sin 37°=2.0\times \sin 37° \\ =1.2\mathrm{m} \end{gathered}$$

∴ Work done when the force is in horizontal direction,

$$\begin{gathered} W\text{'}=mgh \\ =100\times 1.2=120\mathrm{J} \end{gathered}$$