Question:
A block of mass $250 \mathrm{~g}$ slides down an incline of inclination $37^{\circ}$ with a uniform speed. Find the work done against the friction as the block slides through $1 \mathrm{~m}$.
Solution:
Force $=\mu \mathrm{R}$ and $\mathrm{F}=\mathrm{mg} \sin \theta$
$m g \sin \theta=\mu R$
Work done $W=\mu R \cos \theta$
$W=m g \sin \theta \times \cos 0 \times S$
$W=0.25 \times 9.8 \times 0.60 \times 1 \times 1$
$W=1.5 \mathrm{~J}$
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