Q.

Two blocks are connected to an ideal spring of stiffness 200 N/m. At a certain moment, the two blocks are moving in opposite directions with speeds 4 ms1 and 6 ms1, and the instantaneous elongation of the spring is 10 cm. The rate at which the spring energy (kx22) is increasing (in Joules per second) is 100n. Find n.

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answer is 2.

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Detailed Solution

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Increase in potential energy per unit time = decrease in kinetic energy of both

=ddt(12m1v12+12m2v22) =v1(m1dv1dt)+v2(m2dv2dt) =v1(m1a1)+v2(m2a2) Or  dUdt=v1(F1)+v1(F2)     (i) Here,  F1=F2=kx=200 x 0.1=20N

Substituting in Eq. (i) we have,

dUdt=(4)(20)+(6)(20)=200 J/s 100×n=200n=2

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