Question

# A block of mass 5 kg slides down on a rough inclined plane of coefficient of friction 0.75 having angle of inclination 37∘ as shown in the figure. Initially if the block is at rest on the top of inclined plane of length 2 m, then the work done against the friction in moving the block to the bottom of inclined plane is (Take g=10 m/s2).

A
0 J
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B
5 J
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C
10 J
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D
5 J
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Solution

## The correct option is A 0 JGiven, mass of the block (m)=5 kg Length of the inclined plane (s)=2 m By the FBD of the block: R=mgcos37∘ (fs)max=μsR =0.75×mgcos37∘ =34×5×10×45 (fs)max=30 N and the applied force is mgsin37o=5×10×35=30 N Here force due to friction is sufficient to hold the block at rest ∴ Block does not slide down. Hence, displacement =0 ⇒ Workdone against friction =0 Hence option A is the correct answer

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