Question

In one experiment, the students allow the block to oscillate after stretching the spring a distance A. If the potential energy stored in the spring is U0, then what is the change in kinetic energy of the block after it is released from rest and has traveled a distance of A2?

Answer 1

Answer:

    K = m g (A - A2)

Explanation:

In a block spring system the total energy is the sum of the potential energy plus the kinetic energy, for maximum elongation all the energy is potential

         Em = U₀ = m g A

For when the system is at an ele

Elongation A2 less than A, energy has two parts

        Em = K + U₂

       K = Em –U₂

We substitute

     K = m g A - m gA2

    K = m g (A - A2)

Answer 2

Answer:

ΔK.E = U0

Explanation:

Solution:-

- The spring constant = k

- The amount of distance displaced initially xo = A

- The potential energy stored at point xo = U0

- Considering the mass-spring system in isolation there are no external forces like viscous drag or friction acting, then we can safely apply the principle of conservation of energy.

- Which states:

                                       ΔK.E = ΔP.E

Where,  ΔK.E: The change in kinetic energy

             ΔP.E: The change in potential energy

                                      ΔK.E = U0 - Uf

Where,  Uf : Final potential energy.

- The potential energy stored in the spring is given by:

                                     U = 1/2*k*x^2

Where,    x: The displacement of spring from mean position.

- Once the block has been released from displacement x = A/2 about mean position the block travels back to its mean position with displacement x = 0. So the final potential energy when the block has travelled a distance of A/2 is:

                                    Uf = 1/2*k*0^2 = 0

So,

                                   ΔK.E = U0 - Uf

                                   ΔK.E = U0