JEE Main & Advanced
Physics
NLM, Friction, Circular Motion
Question Bank
Kinetic Friction
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 \[={{F}_{k}}\times s=\frac{315.1}{2}\times 2=315\ Joule\]