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 : 

\[\text{ Weight, mg = 100 N } \]
\[\theta = 37^\circ \text{ and s = 2 m }\]
\[\text{ Force, F = mg}  \sin 37^\circ\]
\[ = 100 \times \frac{3}{5} = 60 N\]

So, work done when the force is parallel to incline,

\[W = FS \cos \theta\]
\[ = 60 \times 2 \times \cos 0^\circ= 120 J\]

\[\text{ In } \Delta ABC, AB = 2 \text{ m } \]

\[AC = \text{ h }\]

\[ = s \times \sin 37^\circ= 2 . 0 \times \sin 37^\circ\]

\[ = 1 . 2 \text{ m } \]

∴ Work done when the force is in horizontal direction,

\[\text{ W' = mgh } \]
\[ = 100 \times 1 . 2 = 120 J\]