Question

A 60 kg weight is dragged on a horizontal
surface by a rope upto 2 meter. If coefficient
of friction is u =0.5, the angle of rope with
the surface is 60° and g = 9.8 m/ seca, then
work done is​

Answer 1

Answer :-

$W = 315.2 J$

Given :-

$m= 60 kg \\ s = 2m \\ \mu_k = 0.5 \\ \theta = 60^{\circ}$

To find :-

The work done.

Solution:-

Take block as system.

The force applied on the block is :-

• mg force downward.
• Normal force upward.
• Kinetic friction opposite to the direction of force.
• Components of force in upward and right direction.

See in attachment.

→$N +Fsin60^{\circ} = mg$

→$N = mg - F Cos60^{\circ}$

→$FCos60^{\circ} = F_k$

→$FCos60^{\circ} = \mu_k N$

• Put the value of N.

→$F Cos60^{\circ} = \mu_k ( mg - FSin60^{\circ} )$

• Put the given values.

→$F \times \dfrac{1}{2}= \dfrac{5}{10} ( 60 \times 10 - F \times \dfrac{\sqrt{3}}{2})$

→$\dfrac{F}{2} = \dfrac{1}{2}(600-\dfrac{\sqrt{3}F}{2})$

→ $\dfrac{F}{2}= \dfrac{600}{2}- \dfrac{\sqrt{3}F}{4}$

→$\dfrac{F}{2}+\dfrac{\sqrt{3}F}{4}= 300$

→$\dfrac{2F +F \sqrt{3}}{4} = 300$

→$F(2+\sqrt{3}) = 1200$

→$3.73 F = 1200$

→$F = \dfrac{1200}{3.73}$

→$F = 315 . 1 N$

The friction force will be :-

→$F_k = \dfrac{315.2}{2}$

Work done :-

→$W = F_k\times s$

→$W = \dfrac{315.2}{2}\times 2$

→$W = 315.2 J$

hence,

Work done is 315.2 J.