A) 1m
B) 0.5m
C) 2m
D) 5m
Correct Answer: A
Solution :
[a] Let x be the distance travelled by the particle along the \[37{}^\circ \] incline, after which it comes to rest. From work-energy theorem. Work done by gravity on block on incline \[53{}^\circ +\] work done by friction on block on incline \[53{}^\circ =\] work done by friction on ground level + work done by gravity on block on incline \[37{}^\circ +\] work done by friction on block on incline \[37{}^\circ =\] change in K.E. \[\operatorname{mg}\left( 8 \right)-\mu mg cos 53{}^\circ \left( 10 \right) - umg\left( 10 \right)-\mu mg cos\]\[37{}^\circ \left( x \right)\] \[-mg sin 37{}^\circ \left( x \right) = 0-\frac{1}{2}\left( 1 \right) {{\left( 10 \right)}^{2}}\] On solving we get \[x=1\text{ }m\]
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