# What is the height of the block and how fast is the block moving when the spring reaches its equilibrium position in the following case:A spring with a spring constant of 60 N/m is compressed by...

What is the height of the block and how fast is the block moving when the spring reaches its equilibrium position in the following case:

A spring with a spring constant of 60 N/m is compressed by .05m. The spring is released and a 10kg block sitting on the end of the spring is pushed up a frictionless ramp with an incline of 30 degrees.

*print*Print*list*Cite

### 1 Answer

The work to be done in compressing a spring is dependent on its spring constant and the distance by which it is compressed and is given by (1/2)*k*x^2. In the question you have provided the spring constant as 60 N/m. So to compress it by .05 m requires (1/2)*60*(.05)^2 = .075 J.

The mass placed on the spring has a mass of 10 kg. It is pushed up a ramp inclined at 30 degrees. As the ramp is frictionless, all of the energy in the compressed spring is used to increase the height of the block. The potential energy of the block is given by the product of mass, acceleration due to gravity and the height.

We have m*g*h = .075

=> 10*9.8*h = .075

=> h = .075/98

=> h = .076 cm

At the highest point all the potential energy of the spring has been converted to potential energy stored in the block due to its height.

So, its velocity is zero when the spring reaches the equilibrium position.