A ball with a mass of #256 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #32 (kg)/s^2# and was compressed by #12/4 m# when the ball was released. How high will the ball go?

1 Answer
Apr 17, 2016

Answer:

#57.4m#

Explanation:

Potential energy is stored in a spring as

#P.E.=1/2kx^2#,

where #k# is the spring constant and #x# is how far it has been compressed. Using the values given in the question, #x=12/4m=3m# and #k=32(kg)/s^2#,

#P.E.=1/2*32(kg)/s^2*9m^2#
#=144(kgm^2)/s^2=144Nm#

Due to conservation of energy, we know that the energy before a reaction must be equal to the energy after it, or

#1/2kx^2=1/2mv^2#,

because #1/2mv^2# is the equation for kinetic energy afterwards.

Substituting in values we know already and rearranging to make #v^2# the subject,

#144=1/2*0.256*v^2#
#1125=v^2#

Using the suvat or uvats equation

#v^2=u^2+2as#,

and knowing that #u=0ms^-1#, #a=9.8ms^-2# and #s# is the height (#h#) that we want to find,

#1125m^2/s^2=0m/s+2*9.8m/s^2*hm#
#h=(1125/19.6)m=57.4m#