# How much is the spring compressed by in the following case:A 2 kg disk falls from a height of 20 m and lands on a compressible spring with a spring constant of 10 N/m.

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### 1 Answer

When the disk falls from a height on a spring, the potential energy that was stored in the disk due to its being at a height is converted to potential energy stored in the spring due to it being in a state of compression.

The total energy in the system remains the same (assuming there is no loss due to heat etc.), only the form of the energy changes.

Let us consider the information that is provided. The mass of the disk is 2 kg, its height is 20 m and the spring constant of the spring in 10 N/m.

The potential energy stored in the spring is equal to m*g*h

= 2*9.8*20

= 392

Now assume the disk falls a complete distance of 20 m and the spring is compressed by x m.

The energy stored in the spring is given by (1/2)*k*x^2

So we have (1/2)*k*x^2 = 392

=> (1/2)*10*x^2 = 392

=> x^2 = 392*2/10

=> x = sqrt (78.4)

=> x = 8.85 m

**Therefore the spring is compressed by 8.85 m.**