A) 15J
B) 0.6 J
C) 0.8 J
D) 0.3 J
Correct Answer: B
Solution :
The block comes to rest means its velocity at that point was 3 m/sec. So that point\[K.E.=\frac{1}{2}\times m{{v}^{2}}\] \[=\frac{1}{2}\times 0.1\times {{\left( 3 \right)}^{2}}\] \[=\frac{0.9}{2}=0.45J\]at displacement \[\frac{x}{2}\] \[P.E.=\frac{1}{4}T.E.\] \[K.E.=\frac{3}{4}T.E.\] So \[T.E.=\frac{4}{3}\times 0.45\] \[=0.6J\]You need to login to perform this action.
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