A) 2 J
B) \[3\,J\]
C) \[0\,J\]
D) \[1\,J\]
Correct Answer: D
Solution :
Center of mass of the stick lies at the midpoint and when the stick is displaced through an angle \[60{}^\circ \],it rises up to height h from the initial position. From the figure,\[h=\frac{l}{2}-\frac{l}{2}\cos \theta =\frac{l}{2}(1-\cos \theta )\] Hence, the increase in potential energy of the sick =\[=mgh=mg\frac{l}{2}(1-\cos \theta )=0.4\times 10\frac{1}{2}(1-\cos {{60}^{o}})=1\,J\] Hence, the correction option is (d).You need to login to perform this action.
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