A ball with a mass of #450 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #12 (kg)/s^2# and was compressed by #4/3 m# when the ball was released. How high will the ball go?

1 Answer
May 8, 2017

Answer:

The height is #=2.42m#

Explanation:

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The spring constant is #k=12kgs^-2#

The compression is #x=4/3m#

The potential energy is

#PE=1/2*12*(4/3)^2=32/3J#

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=2/0.45*32/3=47.41#

#u=sqrt47.41=6.89ms^-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=47.41-2*9.8*h#

#h=47.41/(2*9.8)=2.42m#