Question

A body of mass m is pulled along a circular arc by means of a constant horizontal force P as shown. Work done by the force P while lifting the bob from A to B is [l is length of the string and take $sin\left({30}^o\right)$= 0.5]

• A. $mgl$
• B. $\frac{PI}{2}$
• C. $PL$
• D. $\frac{\sqrt{3}}{3}PI$

Answer 1

The correct option is B $\frac{PI}{2}$

Work done, $W=Force\times Displacementinthedirectionofforce\Rightarrow W=P\times BCsin{30}^{\circ }=\frac{BC}{OB}$



In right angled $\Delta OBC,\Rightarrow BC=OBsin{30}^{\circ }\Rightarrow BC=L\left(\frac{1}{2}\right)=\frac{I}{2}(\because OB=I)$
Work done, $W=P\times \left(\frac{I}{2}\right)=\frac{PI}{2}$
Hence, the correct answer is option (b).