A 4.0 kg object is released from rest at a height of 4m on a curved ramp the cur

Question

A 4.0 kg object is released from rest at a height of 4m on a curved ramp the curved surface is frictionless but the horizontal part of it is rough with uk=0.160 the object slides on this horizontal surface for 5.0 m before enocountering the spring with spring consttant k=400 N/m it then compresses the spring with a distance x before momentarily coming to rest a.) determine x b.)will the object ever come to original height explain i would appreciate a detaildd explanation for part a because i tried and got it wrong so i was kind of confused thank you in advance

Explanation / Answer

a) Potential energy of the object at the top of the ramp is converted to its kinetic energy at the bottom of the ramp.

=> mgh = mv12/2

=> v1 = (2gh)1/2 = (2 * 9.81 * 4)1/2 = 8.86 m/s

Speed of the object just before it touches the spring is,

v22 = v12 - 2(kg)s

v2 = [8.862 - (2 * 0.16 * 9.81 * 5)]1/2 = 7.92 m/s

The kinetic energy of the object is converted to the potential energy of the spring.

mv22/2 = kx2/2

=> x = v2(m/k)1/2 = 7.92 * (4 / 400)1/2 = 0.792 m

b) The object will not come back to the same height since its mechanical energy is reduced by frictional force on the surface.

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