Question

Two blocks on a frictionless table are held at rest with a compressed spring between them. When the blocks are released, the spring pushes them apart and they move in opposite directions. The spring was compressed by 0.05 m. One of the blocks moves away at 0.28 m/s while the other moves away at 0.56 m/s. The heavier of the two blocks has a mass of 0.75 kg. What is the spring constant in N/m? (Hint: mechanical energy in this system is conserved)

A 220

B 70.6

C 319

D 12.7

E 44.6

F 56.7

Answer 1

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mo ܓܢ mi m2 let the spring be compressed at x length from both Side between two masses m, & mai In this compressed situation only Potential evingly is stored in the spring When this masses are released, then that .P.E is Converted into Kinetic energies of the two masses. (K. E)masses (p. E.) spreng 2 mivi m2 = 0.75 X0:28 0.56 & kx² = {min? t * ma V2² K(0.05)? (0.75)x(0.28) + m2 10.56)? -0. from linear momentum conservation for two masses mikiamava m = 0.375 kg, but this in K(6.05) ?= (0,0588)+(0.375) x (0.56)2 = 0.0588 +0.1176 K(0.05)} = 0.1764 K= 70.56 N/m ③ obtion is Correct.

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