Question

A man drags a block through $10\, m$ 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. 7.5 kJ
• C. 15 kJ
• D. 10 kJ

Answer 1

Answer: (C)

Horizontal component of applied force
$=\left(\sqrt{3} \times 10^{3}\right) \times \cos 30^{\circ}$
$=\sqrt{3} \times 10^{3} \times \frac{\sqrt{3}}{2}$
$=\frac{3}{2} \times 10^{3} N$
Work done $= F.s$
$=\frac{3}{2} \times 10^{3} \times 10$
$=15 \times 10^{3} J$
$=15\, kJ$