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 \[={{F}_{k}}\times s=\frac{315.1}{2}\times 2=315\ Joule\]You need to login to perform this action.
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