Question

A $60\text{kg}$ weight is dragged on a horizontal surface by a rope upto $2\text{m}$ with constant speed. If the coefficient of friction is $\mu =0.5$, the angle of rope with the surface is ${60}^{\circ }$ and $g=9.8{\text{m/sec}}^2$, then work done by rope on the block is

• A. $294\text{J}$
• B. $315\text{J}$
• C. $588\text{J}$
• D. $197\text{J}$

Answer 1

The correct option is B $315\text{J}$

Let body is dragged with force $P$, making an angle ${60}^{\circ }$ with the horizontal.

Kinetic friction in the motion, $f_k={\mu }_{k}N$
From the figure,
Resolving forces in $x$-axis
$f_k=P\cos {60}^{\circ }$
${\mu }_{k}N=P\cos {60}^{\circ }...\left(1\right)$

Resolving forces in $y$-axis
$N=mg-P\sin {60}^{\circ }$
Equation $\left(1\right)$ becomes
$P\cos {60}^{\circ }={\mu }_k(mg-P\sin {60}^{\circ })$
$\frac{P}{2}=0.5(60\times 9.8-P\times \frac{\sqrt{3}}{2})$
$P=315.1\text{N}$
$\therefore f_k=P\cos {60}^{\circ }$
Work done by rope is $=f_k\times s=315.1\times \frac{1}{2}\times 2\approx 315\text{J}$