A 0.30 kg ladle sliding on a horizontal frictionless surface isattached to one end of a horizontal spring (k = 500 N/m) whoseother end is fixed. The ladle has a kinetic energy of 10 J as itpasses through its equilibrium position (the point at which thespring force is zero). (a) At what rate is the spring doing work on the ladle as the ladlepasses through its equilibrium position? (b) At what rate is the spring doing work on the ladle when thespring is compressed 0.10 m and the ladle is moving away from theequilibrium position?

A 0.30 kg ladle sliding on a horizontal frictionless surface isattached to one end of a horizontal spring (k = 500 N/m) whoseother end is fixed. The ladle has a kinetic energy of 10 J as itpasses through its equilibrium position (the point at which thespring force is zero). (a) At what rate is the spring doing work on the ladle as the ladlepasses through its equilibrium position? (b) At what rate is the spring doing work on the ladle when thespring is compressed 0.10 m and the ladle is moving away from theequilibrium position?

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asked 2021-02-23
A 0.30 kg ladle sliding on a horizontal frictionless surface isattached to one end of a horizontal spring (k = 500 N/m) whoseother end is fixed. The ladle has a kinetic energy of 10 J as itpasses through its equilibrium position (the point at which thespring force is zero).
(a) At what rate is the spring doing work on the ladle as the ladlepasses through its equilibrium position?
(b) At what rate is the spring doing work on the ladle when thespring is compressed 0.10 m and the ladle is moving away from theequilibrium position?

Answers (1)

2021-02-25
(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
\(\displaystyle{P}=-{3.5}\cdot{10}^{{{2}}}{W}\)
0

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