Answer:
The spring constant is k=7031.25
Explanation:
We start by using the conservation of mechanical energy:
Ui + Ki = Uf + Kf
the Ui (initial potential energy) is zero since the spring is relaxed.
In the final situation, the Kf (final kinetic energy is zero since the car and spring are NOT in motion)
Then we have:
[tex]K_i=U_f\\\frac{1}{2} m*v^2=\frac{1}{2} k*(\Delta x)^2\\\frac{1}{2} 2000*15^2=\frac{1}{2} k*(8)^2\\k=7031.25[/tex]
