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