Question

A vertical spring (ignore its mass), whose spring stiffness constant is 880 n/m, is attached to a table and is compressed down 0.170 m. (a) what upward speed can it give to a 0.300 kg ball when released?

Answer 1

When the spring is compressed, the potential energy stored in the spring is:
$U= \frac{1}{2}kx^2$
where k=880 N/m is the spring constant and $x=0.170 m$ is the compression of the spring. By using these numbers, we get
$U= \frac{1}{2}(880 N/m)(0.170 m)^2=12.7 J$

When the spring is released with the ball over it, all the potential energy is converted into kinetic energy of the ball:
$U=K= \frac{1}{2} mv^2$
So by using m=0.300 kg, and re-arranging the formula, we can calculate the velocity of the ball:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2\cdot 12.7 J}{0.300 kg} }=9.2 m/s$