# What is the potential energy stored in the spring when the spring is compressed by 20 cm?

A block of mass 500 grams is attached to a spring with a spring constant k=100 N/m. What is the potential energy stored in the spring when the spring is compressed in the spring when the spring is compressed by 20 cm? If the spring is released after it is compressed, with what velocity will the block leave the spring?

### 5 Answers

- Anonymous1 decade agoFavourite answer
The potential energy stored in a spring is given by this formula:

U = 1/2 • k • x^2

Where U is the potential energy, k is the spring constant, and x is the distance that the spring is compressed or stretched. Thus, in this case:

U = 1/2 • k • x^2

U = 1/2 • 100 N/m • (0.2 m)^2

U = 50 N/m • 0.04 m^2

U = 2 Joules

The kinetic energy of the block will be equal to the potential energy of the spring. This happens because all of the energy stored as elastic potential energy in the spring is converted into kinetic energy when the spring is released. Since KE = 1/2 • m • v^2, we know that:

1/2 • m • v^2 = U

m • v^2 = 2U

v^2 = 2U/m

v = SQRT (2U/m)

By substituting m = 0.5 kg and U = 2 Joules, you get:

v = SQRT (4 J / 0.5 kg)

v = SQRT (8 m^2/s^2)

v = 2.828 m/s

So, our final answer is that the potential energy in the spring is 2 Joules, and the velocity of the block is 2.828 m/s. Hope this helps!

- Anonymous6 years ago
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What is the potential energy stored in the spring when the spring is compressed by 20 cm?

A block of mass 500 grams is attached to a spring with a spring constant k=100 N/m. What is the potential energy stored in the spring when the spring is compressed in the spring when the spring is compressed by 20 cm? If the spring is released after it is compressed, with what velocity will the...

Source(s): potential energy stored spring spring compressed 20 cm: https://tr.im/iqK55 - 1 decade ago
The force in a spring is equal to the spring constant times the displacement of the spring. So, the force F at a displacement of x is:

F(x) = kx

The potential energy is calculated by multiplying the force by the distance moved against the force. The trick with a spring is that the force changes as the displacement changes - it doesn't take much energy to move from 0cm to 1cm, but it takes a lot to move from 19cm to 20cm.

The correct equation is an integral: the total potential energy equals the integral of the force with respect to the displacement, with the limits 0cm to 20cm (0r 0 and 0.2 metres):

PE = {integral from x=0 to x= 0.02}[ F(x)dx ] where F(x) = kx.

Therefore PE = [0.5 * k x^2](0, 0.2) = (0.5 * 100 * 0.2^2) - 0

= 2 joules.

Regarding how fast the block would move when released, you don't say which way teh spring is pointing - so I will assume it is horizontal and we can ignore the effect of gravity. I'm also assuming that the block is on some kind of frictionless trolley, ie all the energy of the spring goes into kinetic energy of the block.

So PE in spring before = KE in block after

2 Joules = 0.5 * mass of block * (velocity squared)

2 = 0.5 * 0.5kg * v^2

v = square root ( 2 /(0.25) ) = square root (8) = 2.8 m/s

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- Anonymous5 years ago
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It is easiest to do the last question first. 1/2 . k . x^2 = energy stored 1/2 . k . 0.02^2 = 10 k = 20 / 0.02^2 = 5 . 10^4 N/m For 4 cm stretch E = 1/2 . 5 . 10^4 . 0.04^2 = 40 J Assuming the spring is hookean,, the energy stored under compression of 2 cm will be the same as that stored under stretching = 10J 20 = 1/2 . 5 . 10^4 . x^2 8 . 10^–4 = x^2 x = 2.83 cm