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 #15 (kg)/s^2# and was compressed by #8/3 m# when the ball was released. How high will the ball go?

Question

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 #15  (kg)/s^2# and was compressed by #8/3 m# when the ball was released.  How high will the ball go?

Amelia Adams · Accepted Answer

The height is #=12.09m#

The spring constant is #k=15kgs^-2#

The compression is #x=8/3m#

The potential energy in the spring is

#PE=1/2*15*(8/3)^2=53.33J#

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#

#PE_(ball)=53.33=0.450*9.8*h#

#h=53.33*1/(0.450*9.8)#

#=12.09m#

Charlotte Aaron · Answer

undefined To find the maximum height the ball will reach, use the conservation of mechanical energy. The potential energy stored in the spring when compressed is equal to the kinetic energy of the ball at its highest point. The potential energy stored in the spring is given by $ PE = \frac{1}{2} k x^2 $, where $ k $ is the spring constant and $ x $ is the compression distance. The kinetic energy of the ball is given by $ KE = \frac{1}{2} mv^2 $, where $ m $ is the mass of the ball and $ v $ is its velocity at the highest point. Setting these equal and solving for $ v $ yields $ v = \sqrt{\frac{kx^2}{m}} $. Then, the maximum height the ball will reach is given by $ h = \frac{v^2}{2g} $, where $ g $ is the acceleration due to gravity. Substituting the given values: $ k = 15 \, (kg)/s^2 $, $ x = \frac{8}{3} \, m $, $ m = 450 \, g = 0.45 \, kg $, and $ g = 9.8 \, m/s^2 $, we can calculate $ v $ and then $ h $. After calculation, the maximum height is approximately $ 0.781 \, m $. Therefore, the ball will go up to a height of approximately $ 0.781 \, m $.