please provide a detailed solution thanks A 0.54 - kg block is held in place aga

Question

please provide a detailed solution thanks

A 0.54 - kg block is held in place against the spring by a 65 - N horizontal external force. The external force is removed, and the block is projected with a velocity v1 = 1.2 m/s upon separation from the spring. The block descends a ramp and has a velocity v2 = 1.4 m/s at the bottom. The track is frictionless between points A and B. The block enters a rough section at B, extending to E. The coefficient of kinetic friction is 0.36. The velocity of the block is v3 = 1.4 m/s at C. The block moves on to D, where it stops. Determine the initial compression of the spring, in cm

Explanation / Answer

k*x = 65N....(1)

(1/2)*m*v^2 = .27 * (1.2)^2 = .389...(2)

also this energy = (1/2) * K * X^2 = 1/2 * 65 * x FROM 1

so equating 1 and 2 we get

x = .012 m or 1.2 cm

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