Question

A spring with a constant of `12 (kg)/s^2` is lying on the ground with one end attached to a wall. An object with a mass of `7 kg` and speed of `3 m/s` collides with and compresses the spring until it stops moving. How much will the spring compress?

Answer 1

By conservation of mechanical energy when the spring is compressed fully after collision
PE gained by the spring = Initial KE of the colliding object
`=>cancel (1/2)kx^2=cancel(1/2)mv^2`
where
m = mass=7kg
v=velocity of the object=`3m/s`
k = force constant=`12(kg)/s^2`
x= compression of spring=?
`=>x^2=mv^2/k`
`=>x=sqrt(m/kv^2)=sqrt(m/k)xxv=sqrt((7kg)/(12kg)/s^2)xx3m/s=2.3m`