How Does Spring Compression Affect Power Transfer to a Moving Ladle?

[Predator]

Homework Statement

A 0.33 kg ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (with k = 485 N/m) whose other end is fixed. The mass has a kinetic energy of 10 J as it passes through its equilibrium position (the point at which the spring force is zero).
(a) At what rate is the spring doing work on the ladle as the ladle passes through its equilibrium position?

(b) At what rate is the spring doing work on the ladle when the spring is compressed 0.10 m and the ladle is moving away from the equilibrium position?

Homework Equations

W = KE(final) - KE(initial)

W = 1/2kx^2

The Attempt at a Solution

Ok, on A, I kinda had to use intuition and guess that when the ladle passes through the equilbrium position, the force is 0, thus the work is 0. Then, that causes me to come to a problem at B, I do not understand how to solve it. I'm guessing I need Force * velocity...and it says that the KE @ equilibrum = 10J, so how do I use that to get V? (K = 1/2mv^2)

 

[Shooting Star]

The total energy, ke+pe, is constant and is given in the problem. It is 10 J. So you know k. From that, find v at 0.10m. The rate of work done is –Fv at a point. F is known for any x.

 

[ogb p]

I would like to clarify a few things about the provided information.

Firstly, it is important to mention that power is the rate at which work is done, and it is measured in units of joules per second (J/s) or watts (W). Therefore, when we are asked to calculate the rate at which work is done, we are essentially being asked to calculate the power.

Secondly, the equations provided in the homework (W = KE(final) - KE(initial) and W = 1/2kx^2) are not directly applicable to this problem. The first equation is used to calculate the work done by a net force on an object, while the second equation is used to calculate the potential energy stored in a spring. In this problem, we need to calculate the work done by the spring on the ladle, which is a different concept.

To solve part (a), we need to use the equation for power, which is P = W/t, where W is the work done and t is the time taken. Since we know the kinetic energy of the ladle (10 J) and the mass of the ladle (0.33 kg), we can use the equation for kinetic energy (KE = 1/2mv^2) to calculate the velocity of the ladle as it passes through the equilibrium position. Once we have the velocity, we can calculate the time taken for the ladle to pass through the equilibrium position (since we know the distance it has to travel). Then, we can plug these values into the equation for power to calculate the rate at which work is done by the spring on the ladle.

To solve part (b), we need to use the same equation for power, but we need to find the work done by the spring on the ladle when it is compressed by 0.10 m. To calculate this, we can use the equation for potential energy stored in a spring (PE = 1/2kx^2) and substitute the given values to find the potential energy stored in the spring. This potential energy is then converted to work done by the spring on the ladle as it moves away from the equilibrium position.

In conclusion, as a scientist, I would approach this problem by clarifying the concepts and equations involved, and then using the appropriate equations to calculate the required values.