Q. A block of mass $1\ kg$ slides down on a rough inclined plane of inclination $60{^\circ}$ starting from its top. If the coefficient of kinetic friction is 0.5 and length of the plane is $1\ m$, then work done against friction is (Take $g = 9.8\ m/s^{2}$)
A) $9.82\ J$
B) $4.94\ J$
C) $2.45\ J$
D) $1.96\ J$
Explanation
(c) $W = \mu\ mg\cos\theta S = 0.5 \times 1 \times 9.8 \times \frac{1}{2} \times 1 = 2.45\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.