Answer
The spring is doing work on the ladle at a rate of $~~-353.5~W$
Work Step by Step
We can find the work the spring does on the ladle as the ladle moves from the equilibrium position to the point where the spring is compressed by $0.10~m$:
$W = -\frac{1}{2}kx^2$
$W = -\frac{1}{2}(500~N/m)(0.10~m)^2$
$W = -2.5~J$
We can find the kinetic energy of the ladle at this point:
$K_f-K_i = W$
$K_f = K_i+W$
$K_f = 10~J-2.5~J$
$K_f = 7.5~J$
We can find the ladle's speed at this point:
$\frac{1}{2}mv^2 = 7.5~J$
$v^2 = \frac{(2)(7.5~J)}{m}$
$v = \sqrt{\frac{(2)(7.5~J)}{0.30~kg}}$
$v = 7.07~m/s$
We can find the spring force at this point:
$F = -kx$
$F = -(500~N/m)(0.10~m)$
$F = -50~N$
We can find the rate at which the spring is doing work on the ladle:
$P = F~v = (-50~N)(7.07~m/s) = -353.5~W$
The spring is doing work on the ladle at a rate of $~~-353.5~W$
