A block of mass $100 \mathrm{~kg}$ slides over a distance of $10 \mathrm{~m}$ on a horizontal surface. If the co-efficient of friction between the surfaces is 0.4, then the work done against friction $(\operatorname{in} J$) is :
Solution
<p>$$\begin{aligned}
& \text { Given } \mathrm{m}=100 \mathrm{~kg} \\
& \mathrm{~s}=10 \mathrm{~m} \\
& \mu=0.4 \\
& \text { As } \mathrm{f}=\mu \mathrm{mg}=0.4 \times 100 \times 10=400 \mathrm{~N} \\
& \text { Now } \mathrm{W}=\mathrm{f} . \mathrm{s}=400 \times 10=4000 \mathrm{~J}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Work, Energy and Power · Topic: Work Done by a Force
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