A block of weight $100 \mathrm{~N}$ is slowly slid up on a smooth incline of inclination $37^{\circ}$ by a person. Calculate the work done by the person in moving the block through a distance of $2.0 \mathrm{~m}$, if the driving force is (a) parallel to the incline and (b) in the horizontal direction.
Force $F=m g \sin \theta$
$F=100 \times \sin 37^{\circ}=60 \mathrm{~N}$
(a) Work done $\mathrm{W}=\mathrm{Fd} \cos \theta$
$W=60 \times 2 \times \cos 0=120 \mathrm{~J}$
(b) In triangle $\mathrm{ABC}, \mathrm{AB}=2 \mathrm{~m}, \Theta=37^{\circ}$
$\mathrm{AC}=\mathrm{h}=1.3$
work done $\mathrm{W}=\mathrm{mgh}$
$W=100 \times 1.2$
$W=120 \mathrm{~J}$
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