Q. A block of mass 10 kg slides down a rough slope which is inclined at $45{^\circ}$ to the horizontal. The coefficient of sliding friction is 0.30. When the block has slide 5 m, the work done on the block by the force of friction is nearly
A) 115 J
B) $75\sqrt{2}\ J$
C) 321.4 J
D) $- 321.4\ J$
Explanation
(b) $f = \mu mg\cos\theta$
or $f = 0.30 \times 10 \times 10\cos{45{^\circ}}$
or $f = \frac{30}{\sqrt{2}}N$
$W = f \times s$
$= \frac{30}{\sqrt{2}} \times 5 = \frac{150}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 75\sqrt{2}\ J$
This is negative work because $f$ and $s$ are oppositely directed
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.