JEE MAIN - Physics (2024 - 29th January Morning Shift - No. 11)
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 :
3900
4500
4200
4000
Explanation
$$\begin{aligned}
& \text { Given } \mathrm{m}=100 \mathrm{~kg} \\
& \mathrm{~s}=10 \mathrm{~m} \\
& \mu=0.4 \\
& \text { As } \mathrm{f}=\mu \mathrm{mg}=0.4 \times 100 \times 10=400 \mathrm{~N} \\
& \text { Now } \mathrm{W}=\mathrm{f} . \mathrm{s}=400 \times 10=4000 \mathrm{~J}
\end{aligned}$$
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