Q. A man drags a block through $10m$ on rough surface $(\mu = 0.5)$. A force of $\sqrt{3}kN$ acting at $30{^\circ}$ to the horizontal. The work done by applied force is
A) Zero
B) $15\ kJ$
C) $5\ kJ$
D) $10\ kJ$
Explanation
(b) [image:5_image213.emf]
The component of applied force $F$ in the direction of motion is $F\cos{30{^\circ}}$. The work done by applied force is
$W = (F\cos{30{^\circ})S = \sqrt{3}} \times 10^{3} \times \frac{\sqrt{3}}{2} \times 10J$
$= 15 \times 10^{3}J = 15kJ$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.