A ball with a mass of #480 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #21 (kg)/s^2# and was compressed by #4/5 m# when the ball was released. How high will the ball go?

Question

A ball with a mass of #480 g# is projected vertically by a spring loaded contraption.  The spring in the contraption has a spring constant of #21  (kg)/s^2# and was compressed by #4/5 m# when the ball was released.  How high will the ball go?

Charles Aeschliman · Accepted Answer

The height is #=1.43m#

The spring constant is #k=21kgs^-2#

The compression is #x=4/5m#

The potential energy in the spring is

#PE=1/2kx^2=1/2*21*(4/5)^2=6.72J#

This potential energy will be converted to kinetic energy when the spring is released and to potential energy of the ball

#KE_(ball)=1/2m u^2#

Let the height of the ball be #=h #

Then ,

The potential energy of the ball is #PE_(ball)=mgh#

The mass of the ball is #=0.480kg#

The acceleration due to gravity is #g=9.8ms^-2#

#PE_(ball)=6.72=0.480*9.8*h#

The height is #h=6.72*1/(0.480*9.8)#

#=1.43m#

Isabella Alvarez · Answer

undefined Use the formula $h = \frac{kx^2}{2mg}$ where $h$ is the height, $k$ is the spring constant, $x$ is the compression distance, $m$ is the mass of the ball, and $g$ is the acceleration due to gravity. Plug in the values to find the height.