A ball with a mass of #420 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #35 (kg)/s^2# and was compressed by #5/4 m# when the ball was released. How high will the ball go?

Question

A ball with a mass of #420 g# is projected vertically by a spring loaded contraption.  The spring in the contraption has a spring constant of #35  (kg)/s^2# and was compressed by #5/4 m# when the ball was released.  How high will the ball go?

Adrian Ayala · Accepted Answer

The height is #=6.64m#

The spring constant is #k=35kgs^-2#

The compression is #x=5/4m#

The potential energy in the spring is

#PE=1/2kx^2=1/2*35*(5/4)^2=27.34375J#

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 #=0.420kg#

#PE_(ball)=24.34375=0.240*9.8*h#

The height is #h=24.34375*1/(0.420*9.8)#

#=6.64m#

Emily Abbate · Answer

undefined Use the energy conservation principle to find the maximum height:

$$ H = \frac{mgh}{k} $$

where
$ m = 0.42 \ \text{kg} $ (mass of the ball),
$ g = 9.8 \ \text{m/s}^2 $ (acceleration due to gravity),
$ h = 5/4 \ \text{m} $ (compression of the spring),
$ k = 35 \ \text{(kg)/s}^2 $ (spring constant).

$$ H = \frac{(0.42)(9.8)(5/4)}{35} $$

$$ H \approx 0.56 \ \text{m} $$