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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 J
Given, 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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