A Block of mass m $$=$$ $$0.1kg$$ 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 the $$distancex2$$ from equilibrium position , it hits another block and comes to rest momentarily, while the other block moves with a velocity $$3m/s$$. The total initial Energy of the spring is,

Question

A Block of mass m $$=$$ $$0.1kg$$  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 the $$distancex2$$   from equilibrium position , it hits another block and comes to rest momentarily, while the other block moves with a velocity $$3m/s$$.   The total initial Energy of the spring is,

Infinity Learn · Accepted Answer

The block comes to rest after collision ⇒block velocity before collision was $$3m/s$$, before collision, $$KE$$ of the spring block system is,  K.E   $$=$$    $$12$$ $$mv2$$  substitute given values ⇒$$KE$$ $$=$$   $$12$$ $$0.1$$×$$32$$ ⇒$$KE$$   $$=$$    $$0.45$$ $$Joule---(1)$$     P.E.   $$=$$ $$12kx2$$ at $$displacement=x2$$ , $$PE$$$$=$$ $$12k(x2)2---(2)$$ Total energy $$=12kA2$$ at displacement x, $$TE$$$$=12kx2---(3)$$ substitute $$equation(3)$$ in equation (2) $$PE$$$$=14TE$$         then remaining energy will be $$KE$$ $$KE$$$$=34TE$$ substitute $$equation(1)$$  $$0.45$$ $$=34TE$$ $$TE$$$$=0.45$$ $$X43$$ Total energy $$=0.6$$ Joule  ∴   T.E   $$=$$    $$43$$×$$0.45$$   $$=$$    $$0.6$$ J

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