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Question
A block of mass 250 g slides down an incline of inclination 37° with uniform speed. Find the work done against friction as the block slides through 1m.
Solution
\[\text{ Given }: \]
\[\text{ m = 250 g,} \theta = 37^\circ, \text{ d = 1 m } \]
Here, R is the normal reaction of the block.
As the block is moving with uniform speed, \[f = \text{ mg } \sin37^\circ\]
So, work done against the force of friction,
\[W = \text{ fd } \cos0^\circ \]
\[W = (\text{ mg } \sin37^\circ) \times d\]
\[W = (0 . 25 \times 9 . 8 \times \sin37^\circ ) \times 1 . 0\]
\[W = 1 . 5 J\]
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