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A elastic spring is compressed by an amount x. Show that its P.E. is 1/2 kx2

where k is the spring constant.

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Answer:

The potential energy (P.E.) stored in an elastic spring is equal to the work done in compressing or stretching the spring.

When a spring is compressed or stretched, the force required to do so is given by Hooke's law, which states that the force F required to compress or stretch a spring by an amount x is given by

F = kx, where k is the spring constant.

The work done in compressing or stretching a spring is equal to the force required to do so multiplied by the distance over which the force is applied.

In the case of a spring, this distance is equal to the amount x by which the spring is compressed or stretched.

Therefore, the work done in compressing or stretching a spring is given by

W = Fd = kx * x = kx^2.

The potential energy stored in the spring is equal to the work done to compress or stretch it.

Therefore, the potential energy of a spring that is compressed by an amount x is given by P.E. = W = kx^2.

Substituting kx^2 = 1/2 kx^2, we find that

P.E. = 1/2 kx^2.

This shows that the potential energy stored in an elastic spring is equal to 1/2 times the spring constant multiplied by the amount x by which the spring is compressed or stretched.

Step-by-step explanation:

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