A particle P of mass m kg is attached to two fixed points A and B by two identic

Question

A particle P of mass m kg is attached to two fixed points A and B by two identical model springs, each of stiffness k and natural length I_0. The point A is at a height 1/4 0 above the point B. The particle is free to oscillate vertically under gravity. The stiffness of each spring is given by k = 4mg/lo. The horizontal level passing through the fixed point A is taken as the datum for the gravitational potential energy. The particle P is initially released from rest at a distance I_o below A. When m = 4.56 kg and l_o = 1.27 m. calculate the total mechanical energy (can be either positive or negative) of the system, in joules. Give your answer to 3 decimal places and take g = 9.81 ms^-2.

Explanation / Answer

WE take Point A as 0 reference potential energy

mass m is lo below point A PE of m   = -mglo

It is released from rest hence KE = 0

spring at A is at its natural length hence PE =0

spring at B is compressed by lo/4

its PE   = +0.5k(lo/4)2   ,   k = 4mg/lo

            = 0.5*(4mg/lo)*(lo/4)2   = mglo/8

Total mechanical energy of the system = -mglo + 0 + mglo/8   = -7mglo/8

                      = -7 *4.56*9.81*1.27/8 = -49.710 J

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