A block of weight $150 \; N$ is pulled $20 \; m$ along a horizontal surface at constant velocity. Calculate the work done by the pulling force if the coefficient of kinetic friction is $0.2$ and the pulling force makes an angle of $60^0$ with the vertical.

Given, (need figure)

Weight of block (W) = mg = 150 N

Workdone $(W) = ?$

The verical and horizontal components of pulling force F are Fcos$\theta$ and Fsin$\theta$. Then,

$R + F.cos60 = mg$

or, R = mg - Fcos60 .......... (i)

        $= 150-\frac{F}{2}$

Also,

$\mu =$$\frac{F.sin60}{R}$

$R = F .$$\frac{\frac{\sqrt{3}}{2}}{0.2}$ .......... (ii)

From equation (i) and (ii), we get

or, $150 - \frac{F}{2} = \frac{\sqrt{3}}{0.4}F$

or, $F = 31.056\;N$

Then work done is,

$W = F.sin60 * 20 = 537.9\;J$

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