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 #6/5 m# when the ball was released. How high will the ball go?

1 Answer
Apr 7, 2017

Answer:

The height is #=3.21m#

Explanation:

enter image source here

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

The compression is #x=6/5m#

The potential energy is

#PE=1/2*21*(6/5)^2=15.12J#

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

#KE=1/2m u^2#

The initial velocity is #=u#

#u^2=2/m*KE=2/m*PE#

#u^2=2/0.48*15.12=63#

#u=sqrt63=7.94ms^-1#

Resolving in the vertical direction #uarr^+#

We apply the equation of motion

#v^2=u^2+2ah#

At the greatest height, #v=0#

and #a=-g#

So,

#0=63-2*9.8*h#

#h=63/(2*9.8)=3.21m#