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Question: A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of \(30 ^ { \circ }\...

A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 30∘30 ^ { \circ } by a force of 10 N parallel to the inclined surface as shown in the figure.

The coefficient of friction between block and the incline is 0.1. If the block is pushed up by 10 m along the incline, then, the work done by applied force is :

(Take g=10 ms−2\mathrm { g } = 10 \mathrm {~ms} ^ { - 2 } )

A

10 J

B

50 J

C

100 J

D

150 J

Answer

100 J

Explanation

Solution

Work done by applied force is

W=Fd=10 N×10 m=100 J\mathrm { W } = \mathrm { Fd } = 10 \mathrm {~N} \times 10 \mathrm {~m} = 100 \mathrm {~J}