A ball with a mass of #40 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #16 (kg)/s^2# and was compressed by #2/5 m# when the ball was released. How high will the ball go?

Question

A ball with a mass of #40 g# is projected vertically by a spring loaded contraption.  The spring in the contraption has a spring constant of #16  (kg)/s^2# and was compressed by #2/5 m# when the ball was released.  How high will the ball go?

Nora Arzate · Accepted Answer

The height reached by the ball is #=3.27m#

The spring constant is #k=16kgs^-2#

The compression of the spring is #x=2/5m#

The potential energy stored in the spring is

#PE=1/2kx^2=1/2*16*(2/5)^2=1.28J#

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=0.040kg#

#PE_(ball)=1.28=0.040*9.8*h#

#h=1.28*1/(0.040*9.8)#

#=3.27m#

The height reached by the ball is #=3.27m#

Olivia Askew · Answer

undefined The height the ball will reach can be calculated using the conservation of energy. The potential energy stored in the compressed spring is converted into the gravitational potential energy of the ball at its highest point.

The potential energy stored in the spring (Us) is given by Hooke's Law: Us = (1/2)kx^2, where k is the spring constant and x is the compression distance.

The gravitational potential energy at the maximum height (Ug) is given by Ug = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height.

Setting these equal and solving for h:
(1/2)kx^2 = mgh

h = (k/m) * x^2/g

Substitute the given values:
h = (16 / (0.04)) * (2/5)^2 / 9.8

Calculate to find the height h.