A spring exerts a force when released of 1000 newtons. The spring is pushing a mass of 3 kilograms. The spring fully extends in 1/50 of a second....

Hello!

Denote the starting force as `F_0,` the mass as `m,` the stiffness of a spring as `k` and the finish time as `t_2.` The starting time is `0.` Also denote the displacement of a mass as `x(t),` let a spring moves to the left and its final position is `0.`

Then we know from Newton's Second law that `F=ma=mx''(t).` Also we know Hooke's law, `F=-kx(t)` (minus sign because `F` acts to the left but...

Hello!

Denote the starting force as `F_0,` the mass as `m,` the stiffness of a spring as `k` and the finish time as `t_2.` The starting time is `0.` Also denote the displacement of a mass as `x(t),` let a spring moves to the left and its final position is `0.`

Then we know from Newton's Second law that `F=ma=mx''(t).` Also we know Hooke's law, `F=-kx(t)` (minus sign because `F` acts to the left but `x(t)` compresses a spring to the right).

So we obtain a differential equation for `x(t),`

`x''=-k/m x.`

And we know `x(t_2)=0` and `x'(t_2)=0` (a mass starts from rest) and `x''(0)=F_0/m.`

The general solution is `x(t)=A sin(t sqrt(k/m)) + B cos(t sqrt(k/m)).`

From the bounding conditions we have `A=0` and `B=F_0/k.`

Because `x(t_2)=0,` we obtain `cos(t_2 sqrt(k/m))=0,` the first time it is when `t_2 sqrt(k/m)=pi/2,` or  `sqrt(k)=(pi sqrt(m))/(2 t_2).`

Our goal is to find `x'(t_2)=-B sqrt(k/m) sin (t_2 sqrt(k/m))=-B sqrt(k/m)=-F_0 /sqrt(km)=-(2F_0 t_2)/(pi m).`

This is the final formula. The magnitude of the velocity in numbers is `(2000/50)/(3.14*3) approx` 4.24 (m/s).