A particle of mass m is attached to one end A of a model spring OA of natural le

Question

A particle of mass m is attached to one end A of a model spring OA of natural length lo and stiffness k. The other end of the spring is attached to a fixed point O. and the spring hangs vertically downwards as shown in Figure Ql. The mass is displaced downwards from its equilibrium position by a distance lo/2 and released from rest. Assume air resistance can be neglected. Figure Q1 Find the expression for the total mechanical energy of the system when the particle is at a distance x below O and is moving with velocity v.

Explanation / Answer

Suppose that the gravitational potential energy is zero at the point O. Therefore gravitational potential energy at a distance x below the O will be mgx. So

Ug = m g x

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Initial length of the spring is l0. Therefore spring potential energy at a distance x below the O will be (1/2)k(l0 - x)2. So

Uspring = (1/2)k(l0 - x)2

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Kinetic energy = K = (1/2) m v2

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Mechanical energy = Kinetic energy + gravitational potential energy + spring potential energy

Mechanical energy = (1/2) m v2 + m g x + (1/2)k(l0 - x)2

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