Answer to Question #114889 in Physics for Bebe

Question

A 5-kg block sits on a frictionless surface, attached to a relaxed spring with a constant of 80 N/m. A string attached to the block runs over a pulley to a 950-g hanging mass. When it is released from rest, how far does the hanging mass fall before rising back up?

Answer 1

The hanging mass and the mass on the table produce tension. They will move with the same acceleration when we release the lower block. Consider the lower block:

$$-mg+T=-ma,\\ T=m(g-a).$$

According to Newton's second law, the acceleration of the block on the horizontal surface can be calculated as

$$-T=-Ma,\\ m(g-a)=Ma,\\ a=g\frac{m}{m+M}.$$

As the string between the mass on the table and the spring stretches, we can write that

$$Ma=kx,\\ x=\frac{Ma}{k}=\frac{gmM}{k(m+M)}=9.8\text{ cm}.$$