A body of mass 10 kg moving with a speed of `2.0 m*s^(-1)` on a frictionless table strikes a mounted spring and comes to rest. If the force constant of the spring be `4xx 10^5 N*m^(-1)` , then what will be the compression on the spring ?

Explanation

The kinetic energy of the body is
`E=1/2 mv^2 ` where m and v be the mass and the speed of the body respectively.
On striking the spring , the kinetic energy of the spring due to compression is completely conveted into the potential energy of the spring . If the spring is compressed through a distance x then its potential energy is
`U=1/2 kx^2`
`therefore 1/2 mv^2=1/2 kx^2`
`or, x=v sqrt(m/k)=2xxsqrt((10)/(4xx10^5)) = 10^(-2) m =1cm`