How Does Spring Force Affect the Kinetic Energy of a Sliding Ladle?

[Theorγ]

Homework Statement

A 0.30 kg ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring with k = 500 N/m, whose other end is fixed. The ladle has a kinetic energy of 10 J as it passes it 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 = F*x
W = (-1/2)kx^2
KE = W

The Attempt at a Solution

My question about this problem is was there an external force that was used to push the ladle away and if this force exists is it used in the calculation of this problem? I haven't be able to really start on it because I haven't been able to decide whether or not this is relevant.

 

[cepheid]

My question about this problem is was there an external force that was used to push the ladle away and if this force exists is it used in the calculation of this problem? I haven't be able to really start on it because I haven't been able to decide whether or not this is relevant.

It's not relevant. You are given the initial conditions, which is that the ladle is at the equilibrium position and is passing across it with a certain non-zero velocity. You don't care how the system got to be in this state, because these initial conditions are enough to predict the subsequent behaviour of the system at any time.

 

[Theorγ]

Okay then that means the velocity at which the ladle is moving with respect to the given kinetic energy and spring force is:

KE = (1/2)mv^2
10 J = (1/2)*0.30kg*v^2
v = 8.2 m/s

KE = -(1/2)kx^2
-10 J = -(1/2)*500 N/m*x^2
x = 0.2 m

x - x0 = v*t
0.2 m - 0 m = 8.2 m/s * t
t = 0.024 s

P = W/t
P = 10 J/0.024 s
P = 410 J/s

Is this correct for question A?

 

[Rayquesto]

I'm just going to throw out my opinion, even though I'm no expert. I'll try to help though. Here's what I find: if the equation for finding work at a position (elastic potential energy) is 1/2 times spring force times distance compressed or stretched squared, then the derivative (rate) is simply the spring constance times the amount of distance stretched or compressed. But 10 joules refers to kinetic energy and the change in initial kinetic energy and initial potential energy is equal to final kinetic energy and final potential energy. So, 10 joules=change in potential energy and 10 joules= 1/2 times spring constance times distance stretched or compressed and if we find that distance, then we can plug that into the derivative (rate) of energy. So, actually the spring has a natural distance .2 meters away from the origin and .2 times 500N/m is 100J/s which could be the rate, but I also could be wrong. I hope that if you know how to do this, can you explain to me what I did wrong.

 

[Theorγ]

I'm just going to throw out my opinion, even though I'm no expert. I'll try to help though. Here's what I find: if the equation for finding work at a position (elastic potential energy) is 1/2 times spring force times distance compressed or stretched squared, then the derivative (rate) is simply the spring constance times the amount of distance stretched or compressed. But 10 joules refers to kinetic energy and the change in initial kinetic energy and initial potential energy is equal to final kinetic energy and final potential energy. So, 10 joules=change in potential energy and 10 joules= 1/2 times spring constance times distance stretched or compressed and if we find that distance, then we can plug that into the derivative (rate) of energy. So, actually the spring has a natural distance .2 meters away from the origin and .2 times 500N/m is 100J/s which could be the rate, but I also could be wrong. I hope that if you know how to do this, can you explain to me what I did wrong.

Hmm well the calculations seem to be right, but what does that mean about the answer I got (410 J/s)?

 

[cepheid]

x - x0 = v*t
0.2 m - 0 m = 8.2 m/s * t
t = 0.024 s

This equation only applies if the object is moving at a constant velocity. It's not, because it has a force acting on it. In fact, it has a continuously changing force acting on it, which means that its acceleration is not even constant.

P = W/t
P = 10 J/0.024 s
P = 410 J/s

Is this correct for question A?

Well, as I pointed out above, your value for t is wrong. I am not sure whether this question is asking you for the power. The "rate at which work is being done" could also mean the rate at which work is being done with distance (i.e. joules/metre). This is just the force.

Is there some way that you can get clarification on what the problem is asking? If you do need to find the power, then you could do it as follows: the rate at which work is being done on the ladle is just the rate of change of its kinetic energy. You know the rate of change of the kinetic energy with distance. In order to find the rate of change with time, you need to know the distance as a function of time. Do you know how to do this?

 

[Theorγ]

Well for one thing I am positive that part A is asking you for the power of the spring.

 

[Rayquesto]

if cephoid is rightabout finding joules per meter being only in force, then .2 meters times 500Newtons per second seems to be right, since were left with only Newtons and meters cancel out.

 

[cepheid]

Sorry, I might have been overcomplicating it. Just use P = F*v.

 

[Rayquesto]

at eq

 

[Rayquesto]

whoops sorry! At equalibrium position, the velocity of the system is at its highest, but net force is zero. idk if this helps. no it doesn't help. um... what other ideas do you have?

 

[cepheid]

whoops sorry! At equalibrium position, the velocity of the system is at its highest, but net force is zero. idk if this helps. no it doesn't help. um... what other ideas do you have?

The way of solving the problem is to do what I said above in post #9. It turns out that

power = force*velocity

The full-on (calculus) derivation of this is that:

W = ∫F(x)dx

Now

P = dW/dt = (dW/dx)*(dx/dt) = F(x)*(dx/dt) = F(x)*v

If you don't understand any of that, don't worry about it. It's just a generalization of the result you get in the specific case (when the force is constant) that since W = FΔx and P = W/Δt, therefore P = F(Δx/Δt) = Fv

 

[Rayquesto]

then isn't power 100Newtons times 8.2meters per second= 820J/s is the answer since the velocity at the equalibrium position is 8.2 meters per second and the force at that point is 100 Newtons?

 

[cepheid]

then isn't power 100Newtons times 8.2meters per second= 820J/s is the answer since the velocity at the equalibrium position is 8.2 meters per second and the force at that point is 100 Newtons?

The force at the equilibrium position is not 100 N. The force at the equilibrium position is 0 N, since F = -kx and x = 0 at the equilibrium position.

Why don't we give the original poster a chance to try it out and report back?

 

[Rayquesto]

haha calculus is always hard for me. were relating it to work now in systems like springs, heights, and other stuff. It makes the stuff we learn in physics pretty complicated, but simplifies things in the end. Well, even though we do the same stuff in physics and that i just use certain energy rules from physics, I seem to get messy doing the physics problems when I have a less understanding in calculus.

 

[Rayquesto]

yup we should. I just don't know how to figure this out any further. it seems that if the force is 0, what does that say about the power?

 

[Theorγ]

Shouldn't the force be 50 N? Since if the work is 10 J, and the distance traveled is 0.2 meters, then 10 J / 0.2 meters = 50 N. And if 50 N is the force then the power is equal to force times velocity, 50 N * 8.2 m/s, which is equal to 410 J/s?

 

[cepheid]

Shouldn't the force be 50 N? Since if the work is 10 J, and the distance traveled is 0.2 meters, then 10 J / 0.2 meters = 50 N. And if 50 N is the force then the power is equal to force times velocity, 50 N * 8.2 m/s, which is equal to 410 J/s?

Unfortunately, this is not correct. If you're considering the displacement from maximum compression to the equilibrium position, then I agree that the the total work done on the object is equal to its change in KE, which is 10 J. However, you CAN'T use F = W/Δx in order to calculate the force. The reason is because this equation is only valid if the force is constant. In this situation, the force is not constant. Remember that, for a spring, the force is given by F = -kx. This is a restoring force that resists any attempt to move the spring away from its equilibrium position (either by stretching or compressing). The force depends on x: the farther you try and stretch or compress it (i.e. the larger x is), the harder it pushes back (the larger F is). Let's choose the direction of compression to be positive, and the direction of stretching to be negative. So, at maximum compression, x = +0.2 m, the spring force is -500 N/m * 0.2 m = -100 N. However, once it has moved half the distance to the equilibrium position, the force has decreased to -500 N/m * 0.1 m = -50 N. By the time it reaches the equilibrium position, the spring force has dropped down to 0 N. So, as you can see, the force is continuously changing throughout the motion during which the 10 J is gained. Although the average force over that distance may be 50 N, that is not what you are interested in. All that matters is the force at the instant that the mass reaches x = 0. At this instant, the force is zero, and the spring is not doing any work on the ladle. Since P = F*v, the power is also 0 at this instant.

I know it must be strange to hear me say that things that you have been taught such as "work = force*distance" are not always true. But what you have to realize is that this relation was derived for the simple case in which you have a constant force. For the more general case, it's more complicated than that. If the force changes over the distance, then which force value do you use when multiplying? Clearly that can't be the whole story. I gave the whole story two posts ago: the most general relation between work and force requires calculus in order to express it. Fortunately, you don't need to resort to using calculus, because P = F*v is always true, even in this more general case of a variable force.

For part b), it's the same procedure, except that you have to use the work-energy theorem in order to figure out what the velocity is at the specified displacement. Say you start at the equilibrium position and are moving inwards (compression). At any given displacement, KE = KEinitial + W. Since W is negative, the KE will be smaller than 10 J for any non-zero x value.