JEE MAIN - Physics (2024 - 30th January Evening Shift - No. 20)
A block of mass $$1 \mathrm{~kg}$$ is pushed up a surface inclined to horizontal at an angle of $$60^{\circ}$$ by a force of $$10 \mathrm{~N}$$ parallel to the inclined surface as shown in figure. When the block is pushed up by $$10 \mathrm{~m}$$ along inclined surface, the work done against frictional force is :
$$\left[g=10 \mathrm{~m} / \mathrm{s}^2\right]$$
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5$$\sqrt3$$ J
5 J
$$5\times10^3$$ J
10 J
Explanation
Work done again frictional force
$$\begin{aligned} & =\mu \mathrm{N} \times 10 \\ & =0.1 \times 5 \times 10=5 \mathrm{~J} \end{aligned}$$
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