Question

A 60 kg weight is dragged on a horizontal surface by a rope up to 2 meters. If coefficient of friction is \[\mu = 0.5,\] the angle of rope with the surface is ${60^0}$ and $g = 9.8m/{\sec ^2}$, then work done is

(A) 294 joules
(B) 315 joules
(C) 588 joules
(D) 197 joules

Answer 1

Hint:Start with finding the force by the rope and then divide it in horizontal and vertical components. The horizontal component so found will be equal to the final force. The find the work done in moving the weight by multiplying the final force and the distance dragged that is 2 m.


Formula used:
Kinetic friction, \[{F_k} = {\mu _k}R\]
Where, ${F_k}$ is the final force.
R is a reaction from the ground surface.
Work done, $W = {F_k} \times S$
Where, S is distance moved.


Complete answer:
Let Force by rope on the body be P.
Angle of rope with body is ${60^0}$. So, there will be vertical and horizontal component of P.
Vertical component = $P\,\sin {60^0}$
Horizontal component = $P\,\cos {60^0}$
Now, the kinetic friction on the body in motion = ${F_k} = {\mu _k}R$
Final force will be equal to the horizontal component of P.
${F_k} = P\cos {60^0}$(equivalent horizontal force) equation1
And the reaction force, $R = mg - P\sin {60^0}$(resultant vertical force)
Also, ${F_k} = {\mu _k}R = {\mu _k}(mg - P\sin {60^0})$ equation2
From equation 1 and 2, we get;
$P\cos {60^0} = {\mu _k}(mg - P\sin {60^0})$
Putting all the values here;
$P \times \dfrac{1}{2} = 0.5(60 \times 10 - P \times \dfrac{{\sqrt 3 }}{2})$
Further solving, we get;
P = 315.1 N
Now, work done $W = {F_k}S$
$W = P\cos {60^0} \times 2 = \dfrac{{315.1}}{2} \times 2$
$W = 315.1N \approx 315N$

Hence, the correct answer is Option(B).




Note: Be careful about the vertical and horizontal component of the forces applied and find the resultant force in vertical and horizontal direction accordingly. Also the angle of depression will be the same for both components. Here it was 60 degrees, it will be different in different questions and the answer will change accordingly.