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CaHeK987 [17]
3 years ago
15

In one experiment, the students allow the block to oscillate after stretching the spring a distance A. If the potential energy s

tored 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?
Physics
2 answers:
atroni [7]3 years ago
3 0

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)

victus00 [196]3 years ago
3 0

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

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