Question

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?

Answer 1

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`