A ball with a mass of #350 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 #350 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?

Charles Aeschliman · Accepted Answer

The height is #=7.97m#

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

The compression is #x=5/4m#

The potential energy in the spring is

#PE=1/2*35*(5/4)^2=27.34J#

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 #

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

Then ,

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

Mass of the ball is #m=0.350kg#

#PE_(ball)=27.34=0.35.*9.8*h#

#h=27.34*1/(0.35*9.8)#

#=7.97m#

The height is #=7.97m#

Leo Abeyta · Answer

undefined To find the maximum height reached by the ball, you can use the principle of conservation of mechanical energy. At the highest point, all the initial kinetic energy of the ball will be converted into gravitational potential energy. The equation for gravitational potential energy is $mgh$, where $m$ is the mass of the ball, $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$), and $h$ is the maximum height.

The initial kinetic energy of the ball is equal to the energy stored in the compressed spring, which is given by $ \frac{1}{2} kx^2$, where $k$ is the spring constant and $x$ is the compression of the spring.

Using these equations, you can find the maximum height $h$ reached by the ball.