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A $2 \mathrm{~kg}$ mass lying on a table is displaced in the horizontal direction through $50 \mathrm{~cm}$. The work done by the normal reaction will be
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Since, displacement on the body of mass $2 \mathrm{~kg}$ is in the horizontal direction while normal reaction acts on the body in perpendicular to the displacement.
Hence, $\quad \theta=0$
$\therefore$ Work done, $W=F s \cos 90^{\circ}=0$
Hence, $\quad \theta=0$
$\therefore$ Work done, $W=F s \cos 90^{\circ}=0$
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