Work Energy Power

JEE Main level, single correct, medium

12 of 183

Question 12 of 183

A block of mass 100kg100kg slides over a distance of 10m10m on a horizontal surface. If the co-efficient of friction between the surfaces is 0.40.4 , then the work done against friction (in JJ) is:

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Answer: (C) 40004000

Given m=100kgm=100kg
s=10ms=10m
μ=0.4\mu =0.4
The frictional force is given by
f=μmg=0.4×100×10=400N\begin{matrix}f & = & \mu mg \\ & = & 0.4\times 100\times 10 \\ & = & 400N\end{matrix}
So, the work done by the force can be calculated as follows:
W=f×s=400×10=4000J\begin{matrix}W & = & f\times s \\ & = & 400\times 10 \\ & = & 4000J\end{matrix}

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