Question

A man drags a block through $10m$ on rough surface ($\mu =0.5$). Force of $\sqrt{3}$kN acting at ${30}^o$ to the horizontal. The work done by the applied force is:

• A. $Zero$
• B. $15kJ$
• C. $5kJ$
• D. $10kJ$

Answer 1

The correct option is B $15kJ$

Horizontal component of applied force =$(\sqrt{3}\times {10}^3)\times \cos {30}^o$

=$\sqrt{3}\times {10}^3\times \frac{\sqrt{3}}{2}$

=$\frac{3}{2}\times {10}^{3}N$

Work done = $F.s$

=$\frac{3}{2}\times {10}^3\times 10$

=$15\times {10}^{3}J$

=$15kJ$