A ball with a mass of 400 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 25 (kg)/s^2 and was compressed by 3/7 m when the ball was released. How high will the ball go?

1 Answer
Sep 26, 2017

Potential Energy stored in spring if it compressed by x meter
U=1/2 kx^2
where k="spring constant"
x="Compression"
m=0.4 kg
U=1/2(25)(3/7)^2=(25xx9)/(2xx49)

if spring release the ball will get kinetic energy which is equal to the potential energy of spring. (Conservation of energy)
K=1/2mv^2=(25xx9)/(2xx49)
=>0.4v^2=(25xx9)/(49)
=>v^2=(25xx9)/(0.4xx49)=(25xx9xx10)/(4xx49)
=>v=sqrt((25xx9xx10)/(4xx49)
=>v=(5xx3xxsqrt10)/(2xx7)=15/14sqrt10

To find the height we use Newton's Equation
v^2=u^2+2as
where u=initial velocity =15/14sqrt10
v="final velocity"=0
a=-g="acceleration due to gravity"=-9.8m/s^2

Hence 0^2=(15/14)^2xx10-2(9.8)s
s=(15/14)^2(10/(2xx9.8))=2250/3841.6=0.585 m