What is the Work Done in Solving a Force and Energy Problem?

[MellowOne]

[SOLVED] Work and Energy Problem

Homework Statement

A force is applied to a rigid body moving it from rest at point (1.00 m, 4.00 m) to rest at point (5.00 m, 8.00 m). The force has an x-component of (3.00 N/m^2)x^2 and a y-component of (3.95 N/m)y. Determine the work done.

Homework Equations

KE = 1/2mv^2
PE = mgh
J = kgm^2/s^2
N = kgm/s^2

The Attempt at a Solution

I have it drawn out on a graph. One arrow pointing to the right is the "x-component" of (3.00N/m^2)x^2 and a "y-component" of (3.95N/m)y. I know I have to get the answer in Joules...which is J = kgm^2/s^2. Not sure what to do because the x^2 and y after each component is throwing me off. Any idea?

 

[dynamicsolo]

Not sure what to do because the x^2 and y after each component is throwing me off. Any idea?

What this means is that the force components are not constant, but are functions of the position of the particle. That is to say, the components of the force at the initial position are

Fx = 3 · 1^2 = 3 N , Fy = 3.95 · 4 = 15.8 N

and at the final position are

Fx = 3 · 5^2 = 75 N , Fy = 3.95 · 8 = 31.6 N .

So you will need to do an integration along the path of the particle in order to find the total work done. Have you dealt with work as an integral of the dot product of force and infinitesimal displacement (integral of F · dx)?

 

[MellowOne]

What this means is that the force components are not constant, but are functions of the position of the particle. That is to say, the components of the force at the initial position are

Fx = 3 · 1^2 = 3 N , Fy = 3.95 · 4 = 15.8 N

and at the final position are

Fx = 3 · 5^2 = 75 N , Fy = 3.95 · 8 = 31.6 N .

So you will need to do an integration along the path of the particle in order to find the total work done. Have you dealt with work as an integral of the dot product of force and infinitesimal displacement (integral of F · dx)?

Oh, Ok. I get what you did for finding the N for each part. But integration part I'm totally lost. To answer your question, no we havn't. Is there another way to do this problem?

 

[dynamicsolo]

Have you dealt with integration to find work done along a path at all? If so, there's a way to set it up. There isn't really any good way to do this otherwise when the force is not a constant vector over the entire path.

 

[MellowOne]

Have you dealt with integration to find work done along a path at all? If so, there's a way to set it up. There isn't really any good way to do this otherwise when the force is not a constant vector over the entire path.

No we havn't done any integration. Is it possible to guide me through it or "teach" me it?

 

[dynamicsolo]

First off, is your course calculus-based at all? I'm wondering if they're going to show you how to do this sort of problem...

We know that the force is described by Fx = 3·(x^2) , Fy = 3.95·y. What is the equation of the line connecting the starting and ending points?

 

[MellowOne]

First off, is your course calculus-based at all? I'm wondering if they're going to show you how to do this sort of problem...

We know that the force is described by Fx = 3·(x^2) , Fy = 3.95·y. What is the equation of the line connecting the starting and ending points?

There's suppose to be a little bit of calculus. The thing is I didn't take calculus yet so I don't know some of the math other's in the class might.

line - y = x + 3

(8-4)/(5-1) = slope = 1
8 = 1(5) + b
b = 3

 

[dynamicsolo]

Arrggh! I'm really sorry: I just re-read the problem and realized what they actually asked you to do! Most of the information in this problem is window-dressing or distraction. You are told that the object is at rest at the beginning and at rest at the end. (I don't know if they've talked to you about how to test for a conservative force.) Have you had the work-kinetic energy theorem yet? (I presume by "determine the work done" that the problem is asking for the net work done on the object.)

 

[MellowOne]

Arrggh! I'm really sorry: I just re-read the problem and realized what they actually asked you to do! Most of the information in this problem is window-dressing or distraction. You are told that the object is at rest at the beginning and at rest at the end. (I don't know if they've talked to you about how to test for a conservative force.) Have you had the work-kinetic energy theorem yet?

Is that KEf - KEi = Total Work?

 

[dynamicsolo]

Is that KEf - KEi = Total Work?

So the total work done on the object is...?

I'm sorry about that -- there are many levels of physics courses dealt with in this Forum. It would be a possible question to ask what the work done by the force is on the object in the situation you described in the problem. But when you said you hadn't done much with integration, I was puzzled as to how they expect you to solve the "path integral" (as it's called) for the work without using calculus (it could be managed if both components had linear functions, but not with a quadratic like 3·(x^2)...)

 

[MellowOne]

So the total work done on the object is...?

I'm sorry about that -- there are many levels of physics courses dealt with in this Forum. It would be a possible question to ask what the work done by the force is on the object in the situation you described in the problem. But when you said you hadn't done much with integration, I was puzzled as to how they expect you to solve the "path integral" (as it's called) for the work without using calculus (it could be managed if both components had linear functions, but not with a quadratic like 3·(x^2)...)

KEf = KEi

It wouldn't be 0 would it?

Edit: Cause work HAD to be done to move the body.

 

[dynamicsolo]

KEf = KEi

It wouldn't be 0 would it?

Edit: Cause work HAD to be done to move the body.

Here's the thing about this problem. The force described in the problem would do positive work on the object, so it would arrive at the endpoint with a non-zero kinetic energy. The fact that it didn't means there must have been at least one other force doing negative work on it as well. In the context of the problem, the question would seem to be asking for the net work on the body, which would be zero.

This seems to be the only way the problem makes sense, given what you've been taught to do so far.

 

[MellowOne]

Here's the thing about this problem. The force described in the problem would do positive work on the object, so it would arrive at the endpoint with a non-zero kinetic energy. The fact that it didn't means there must have been at least one other force doing negative work on it as well. In the context of the problem, the question would seem to be asking for the net work on the body, which would be zero.

This seems to be the only way the problem makes sense, given what you've been taught to do so far.

Well I tried using 0 as an answer but it wasn't right. Here's my new thought process. What if we found the distance it traveled using what we know? It traveled 4 m to the right and 4 m up. So using a^+b^2 =c^2 we can find the distance which was 5.65685m. Now this is the part I'm stuck. Since W = F x d, we need F right? How would we go about finding F. Using what you said before there might be a way to incorporate this?

 

[dynamicsolo]

All right, so they do want the work done by the force. Then we have to do the integration. Neither the magnitude nor the direction of the force is constant over the path y = x+3 that we are following from x = 1 to x = 5 , so there is no simple formula for this.

We already found the equation for the line along which the object travels. The components of the force are

Fx = 3·(x^2) and Fy = 3.95·y = 3.95·(x+3) ,

where we make the substitution from the path equation in order to eliminate y. The infinitesimal work done on an infinitesimal bit of the path at (x,y) is then

dW = Fx · dx + Fy · dy .

The differential step we need for dy is given by

dy = (dy/dx) · dx = [ d/dx (x+3) ] · dx = 1 · dx = dx ,

so the infinitesimal work is

dW = 3·(x^2)·dx + 3.95·(x+3)·dx = [ 3·(x^2) + 3.95·(x+3) ]·dx ,

which we must now integrate from x = 1 (corresponding to the point (1,4) )
to x = 5 (corresponding to (5,8) ).

Do you understand how to do this part?

 

[MellowOne]

All right, so they do want the work done by the force. Then we have to do the integration. Neither the magnitude nor the direction of the force is constant over the path y = x+3 that we are following from x = 1 to x = 5 , so there is no simple formula for this.

We already found the equation for the line along which the object travels. The components of the force are

Fx = 3·(x^2) and Fy = 3.95·y = 3.95·(x+3) ,

where we make the substitution from the path equation in order to eliminate y. The infinitesimal work done on an infinitesimal bit of the path at (x,y) is then

dW = Fx · dx + Fy · dy .

The differential step we need for dy is given by

dy = (dy/dx) · dx = [ d/dx (x+3) ] · dx = 1 · dx = dx ,

so the infinitesimal work is

dW = 3·(x^2)·dx + 3.95·(x+3)·dx = [ 3·(x^2) + 3.95·(x+3) ]·dx ,

which we must now integrate from x = 1 (corresponding to the point (1,4) )
to x = 5 (corresponding to (5,8) ).

Do you understand how to do this part?

I actually figured out the question already. Thanks for the help though. Looking at what you did before I got through most of it. The only part that still puzzles me is why the distance you use to get the correct answer is 4m when in reality the answer should be found using 5.65m which is from using the Pythagorean theorem.

 

[dynamicsolo]

The only part that still puzzles me is why the distance you use to get the correct answer is 4m when in reality the answer should be found using 5.65m which is from using the Pythagorean theorem.

It's true that the total path length is 4·sqrt*(2) meters. Both components of the force are doing work along the path simultaneously, of course, but the computation of the amount of work done by each component can be done separately and added together at the end. Because we have advanced 4 meters in the x- and y- directions at the same time, the total path length is covered. (We eliminated y in the integration because it is possible to do so and makes the calculation easier: we only need to use calculus of one variable, rather than of two.)