(a)Since the force exertedby the spring on the mass 0.30 kg is zero when the mass passesthrough the equilibrium position of the spring, the rate at whichthe spring is doing work on the mass 0.30 kg at this instant isalso zero.

(b)Force (F) = k x

= 500 * 0.1 N

= 50.0 N

Apply, Law of conservation of energy:

\(\displaystyle{10.0}{J}={\left(\frac{{1}}{{2}}\right)}{m}{v}^{{{2}}}+{\left(\frac{{1}}{{2}}\right)}{k}{x}^{{{2}}}\)

or v = 7.1 m/s

rate is the spring doing work on the ladle when the springis compressed 0.10m

P = - F . v

= 50.0 * 7.1 W

\(\displaystyle{P}=-{3.5}\cdot{10}^{{{2}}}{W}\)

(b)Force (F) = k x

= 500 * 0.1 N

= 50.0 N

Apply, Law of conservation of energy:

\(\displaystyle{10.0}{J}={\left(\frac{{1}}{{2}}\right)}{m}{v}^{{{2}}}+{\left(\frac{{1}}{{2}}\right)}{k}{x}^{{{2}}}\)

or v = 7.1 m/s

rate is the spring doing work on the ladle when the springis compressed 0.10m

P = - F . v

= 50.0 * 7.1 W

\(\displaystyle{P}=-{3.5}\cdot{10}^{{{2}}}{W}\)