$300 J$ of work is done is sliding a $2 k g$ back up on inclined plane of height $10 m$ . Taking $g = 10 m / s^{2}$ , work done against friction is:

200 J

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100 J

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Z e r o

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1000 J

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Solution

The correct option is B $100 J$
Given that,

Actual work done $W = 300 J$

Mass of block $m = 2 k g$

Inclined plan of height $h = 10 m$

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

$W = W_{g} + W_{f}$

$300 - m g h = W_{f}$

$W_{f} = 300 - 2 \times 10 \times 10$

$W_{f} = 100 J$

Hence, the work done against friction $100 J$

Suggest Corrections

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