Question

A block of mass 20 kg is slowly slid up on a smooth incline of inclination${53}^o$ by a person. Calculate the work done by the person in moving the block through a distance 4 m, if the driving force is (a) parallel to the incline and (b) in the horizontal direction. $[g=10m/s^2]$

Answer 1

(a)The work done =$WD=\overrightarrow{F}\cdot \overrightarrow{d}$

Since the force$\overrightarrow{F}$ and displacement $\overrightarrow{d}$ are in the same direction.

$WD=Fd$

Here

$F=mgsin{53}^0=20\cdot 10\cdot sin{53}^0=200\cdot 0.7986=159.727N$

Hence the work done

$F\cdot d=159.727\cdot 4=638.9J$

(b)

From the figure (b)

As the block is in equilibrium.

Net force along the plane=0.

$F_{1}cos53^0=mgsin53^0{F}_1=mg\cdot tan53^0$

The force parallel to plane=$F_{1}cos53^0=mgsin53^0$

Therefore,

$WD=F_1\cdot d=mgsin53^0\cdot 4=159.727\times 4=638.9J$