Question

A 0.500-kg object is connected to a horizontal massless spring. The spring is initially stretched by 0.200 m and the object is released from rest there. It proceeds to move without friction. The next time the speed of the object is zero is 0.500 s later. Determine the maximum speed of the object, the spring constant, and the mechanical energy of the system. Determine potential energy of the object when it’s speed is a third of the maximum speed.

Answer 1

period of the motion is T = 2*0.5 = 1 s

angualr frequency is w = 2*pi/T

W = 2*3.142/ 1= 6.28 rad/s

Amplitude is A = 0.2 m

maximum speed is vmax = A*w = 0.2*6.28 = 1.256 m/s

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PE = mgh

When V = 1/3 of 1.256 m/s = 0.418 m/s

max height H = v^2/2g = 0.418*0.418/(2* 9.8)

H = 0.0089 m

energy Potneitial = mgh = 0.5 * 9.8 * 0.0089

PE = 0.04361 J