question E Two blocks of masses m and 3m are placed on a frictionless, horizontal surface. A light spring is attached to the more massiveblock, and the blocks are pushed together with the spring between them as shown in the figure below. A cord initially holding theblocks together is burned; after that happens, the block of mass 3m moves to the right with a speed of v.1.65 m/s.3mTwo images, figure a and figureb, depict two differentconfiguransNA ocm assm, a block omens sm, a cord,and a spring.a. Before: Rcoock of massm on the let is connected toa he block of mass 3m on theright by a taut cord and by aVm pressed spring. V3mArter: The block fmass mis onthe le nd the bidckof mass 3mis on tmight.m.2he ered and theCospring, no fongercompressed, is attached onits rightAfterhe block ofbmass 3m, but its left end isunattached. Vector v is(a) What is the velocity of the block of mass m? (Assume right is positive and left is negative.)-4.9m/s(b) Find the system's original elastic potential energy, taking m = 0.470 kg.7.7(c) Is the original energy in the spring or in the cord?in the springin the cord(d) Is the momentum of the system conserved in the bursting-apart process?YesNo(e) If the spring was compressed by 5.30 cm, what is the spring constant of the spring?X N/m

Conservation of Energy For an isolated system, the initial total energy of the system is equal to the final total energy. Initial Total Energy = Final Total Energy If the spring constant of a spring is k then the potential energy stored in the spring for x distance compression is given by US=12kx2We know that the total energy is conserved for the above system of blocks. Therefore in the first figure, it is provided that the compression of the spring is x=5.3 cm. If k is the spring constant of the spring then the potential energy of the spring is Us=12kx2=12k×5.31002 You have found that the velocity of the block of mass m is v1=4.9 m/s and that of the block of mass 3m is v2=1.65 m/s. Therefore the total energy of the system after the spring releases is purely kinetic TE=12×m×4.92+12×3m×1.652 The value of m given in the question is m=0.47 kg. Therefore the total energy after spring release is TE=12×0.47×4.92+12×3×0.47×1.652=7.56 J Therefore by conservation of total energy we have 12k×5.31002=7.56 J Therefore the spring constant is given by k=2×7.56×1005.32=5383.9 N/mAnswer: The value of the spring constant is 5383.9 N/m.

Answered: (e) If the spring was compressed by…

Science

Advanced Physics

(e) If the spring was compressed by 5.30 cm, what is the spring constant of the spring? N/m