A ball with a mass of #140 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #21 (kg)/s^2# and was compressed by #3/8 m# when the ball was released. How high will the ball go?

1 Answer
Oct 21, 2017

Answer:

1.076 m

Explanation:

Potential stored in spring #= (kx^2)/2 = (21 kg.s^-2 × (3/8 m)^2)/2 = 189/128J#

Kinetic energy gained by ball #=(mv^2)/2#
#= (140 cancel(g) × 10^-3 (kg)/cancel(g) × v^2)/2 = (0.07 kg) × v^2#

According to law of conservation of energy
Potential energy in spring = Kinetic energy gained by ball

#189/128 J = (0.07 kg) × v^2#

#v^2 = 21.09375 J/(kg)#

#H = v^2/(2g) = (21.09375 J/(kg)) / (2 × 9.8 ms^-2)= 1.076 m#