question_answer

A block of mass m = 0.1 kg is connected to a spring of unknown spring constant k. It is compressed to a distance x from its equilibrium position and released from rest. After approaching half the distance \[\left( \frac{x}{2} \right)\]from equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with a velocity \[3m{{s}^{-1}}.\]The total initial energy of the spring is: JEE Main Online Paper (Held On 10 April 2015)

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 \[T.E.=\frac{4}{3}\times 0.45\]             \[=0.6J\]