Question

A plane surface is inclined at an angle of ${60}^0$ with the horizontal. A body of mass $10$ kg is uniformly accelerating, along the inclined plane surface. The value of coefficient of friction ${\mu }_k$, between the body and the inclined surface is $0.2$, if the length of the inclined plane is $10$ m, then the work done to pull it to the top is ( Take $g=10m/s^2$)

• A. 666 J
• B. 766 J
• C. 866 J
• D. 966 J

Answer 1

Answer: (D)

966 J

F $=mg(\sin \theta +{\mu }_k\cos \theta )$

$=10\times 10\times (\frac{\sqrt{3}}{2}+0.2\times \frac{1}{2})$

$=100\left(0.966\right)=96.6N$

Work Done $=F.s$

$=96.6\times 10=966J$