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Question

300 J of work is done is sliding a 2 kg back up on inclined plane of height 10 m. Taking g=10 m/s2, work done against friction is:

A
200 J
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B
100 J
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C
Zero
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D
1000 J
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Solution

The correct option is B 100 J

Given that,

Actual work done W=300J

Mass of block m=2kg

Inclined plan of height h=10m

We know that,

Net work done sliding a block up to a height h on inclined plane = work done against gravitational force +work done against frictional force

W=Wg+Wf

300mgh=Wf

Wf=3002×10×10

Wf=100J

Hence, the work done against friction 100 J


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