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 ONLINE 10-04-2015]

A) 15 J

B) 0.6 J

C) 0.8 J    

D) 0.3 J

Correct Answer: B

Solution :

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