A ball with a mass of #144# #g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #48# # kgs^-2# and was compressed by #8/3# #m# when the ball was released. How high will the ball go?

1 Answer
Jul 11, 2016

Answer:

#E_"sp"=1/2kx^2=1/2xx48xx(8/3)^2=170.7# #J#

#E_"gp"=mgh# and, rearranging to make the height the subject, #h=E_"gp"/(mg)=170.7/(0.144xx9.8)=121# #m#

Explanation:

The spring potential energy is given by #E_(sp)=1/2kx^2=1/2xx48xx(8/3)^2=170.7# #J# where #k# is the spring constant and #x# is the distance the spring is compressed.

The gravitational potential energy is given by #E_gp=mgh#, and we know it will be equal to the spring potential energy, because in this case one form of energy is being converted into another. Spring potential energy becomes kinetic energy becomes gravitational potential energy.

Rearrranging:

#h=E_"gp"/(mg)=170.7/(0.144xx9.8)=121# #m#

Note that I remembered to convert the mass of #144# #g# to the correct SI units, #0.144# #kg#.