question_answer

A 60 kg weight is dragged on a horizontal surface by a rope upto 2 metres. If coefficient of friction is \[\mu =0.5\], the angle of rope with the surface is 60° and \[g=9.8\,m/{{\sec }^{2}}\], then work done is                                    [MP PET 1995]

A)                         294 joules                 

B)                         315 joules            

C)                         588 joules                 

D)                           197 joules

Correct Answer: B

Solution :

                            Let body is dragged with force P, making an angle 60° with the horizontal.             \[{{F}_{k}}=\] Kinetic friction in the motion =\[{{\mu }_{k}}R\]             From the figure \[{{F}_{k}}=P\cos 60{}^\circ \ \] and \[R=mg-P\sin 60{}^\circ \]             \[\therefore \]\[P\cos 60{}^\circ ={{\mu }_{k}}(mg-P\sin 60{}^\circ )\]             Þ \[\frac{P}{2}=0.5\left( 60\times 10-\frac{P\sqrt{3}}{2} \right)\] Þ \[P=315.1\ N\]             \[\therefore \] \[{{F}_{k}}=P\cos 60{}^\circ =\frac{315.1}{2}N\]             Work done \[={{F}_{k}}\times s=\frac{315.1}{2}\times 2=315\ Joule\]