Answer to Question #221965 in Physics for caleb villarreal

Question

A spring with a constant k= 200N/m is attached to a wall horizontally as shown. It is stretched a distance of x = 0.25m. The block attached has a mass of 2kg. The block is released and allowed to move. Assume the surface is frictionless and the block-spring is at ground level.  

 

1. Find the velocity of the block at the moment it reaches the springs equilibrium point. 
2. How far past the equilibrium point will the block compress the spring?  

Answer 1

1. At the moment the block reaches the spring's equilibrium point, the potential energy of the spring converts into kinetic energy of the block:

$$E_p=E_k,\\\space\\ \frac12 kx^2=\frac12mv^2,\\\space\\ v=x\sqrt{\frac km}=2.5\text{ m/s}.$$

2. Assuming no energy was lost, the elongation of the spring will be 0.25 m from the equilibrium point.