Answer
a) $W_g=258,000J$
b) $v=23.6\frac{m}{s}$
c) $x=-2.62m$
Work Step by Step
a) $W_g=\Delta E_P$
$\Delta E_{Pf}-\Delta E_{Pi}=mg(h_f-h_i)=(925kg)(9.8\frac{kgm}{s^2})(28.5m)=258,353J=258,000J$
b) The potential energy is now 0 so only kinetic energy is left
$\frac{mv^2}{2}=258,353J$
$v=\sqrt{\frac{2\times258,353J}{925kg}}=23.6\frac{m}{s}$
c) $258,353J=\frac{kx^2}{2}+mgx=\frac{(8.00\times10^4\frac{N}{m})x^2}{2}+(625kg)(9.8\frac{m}{s^2})x=4.00\times10^4\frac{N}{m}x^2+6125Nx$
Solving the quadratic equation, the possible answers for x are: $-2.62$ and $2.47$. Because the spring is compressed down, the value of x will be negative so $x=-2.62m$