# A mass of 10 kg is dropped on a spring 1 m long kept on the ground from a height of 5 m. If the spring constant is 1280 N/m how much is the spring compressed.

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### 1 Answer

The spring constant of the spring is 1280 N/m^2. It is 1 m long and kept on the ground. Let the spring be compressed by a length x m when the mass of 10 kg is dropped on it from a height of 5 m.

The gravitational potential energy of the mass at a height of 5 m is m*g*h = 10*5*9.8 = 490 J. When it falls on the spring, the gravitational potential energy is equal to 10*9.8*(1 - x) and the potential energy in the spring is equal to (1/2)*k*x^2 = (1/2)*1280*x^2. As the total energy in the system is conserved it gives:

490 = 10*9.8*(1 - x) + (1/2)*1280*x^2

=> 640*x^2 - 98*x - 392 = 0

Solving the quadratic equation gives a positive solution approximately equal to 0.8629 m.

(The negative solution can be ignored as the spring is being compressed not stretched.)

When the mass is dropped on the spring it is compressed by a length equal to 0.8629 m

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