# Work-Energy Theorem locomotive Question

1. Dec 25, 2013

### Speedking96

1. The problem statement, all variables and given/known data

A locomotive exerts a 20 000 Newton force upon a train. It propels four cards, each with a 15 tonne mass. The locomotive's mass is 40 tonnes and friction is considered negligible. Initially at rest, the train accelerates over a distance of one kilometer.

a) What is the work done by the locomotive on the train?
b) What is the total work?
c) What is the velocity at the end of the 1 kilometer?

2. Relevant equations

F=ma

Ek = (m v^2)/2

W = F * d

3. The attempt at a solution

(a)

w = f * d = (20 000)(1000) = 20 000 000 Joules

(b) How is that different from question (a)?

(c) 20 000 000 = 0.5 * 100 000 * v^2

v = 20 m/s

2. Dec 25, 2013

### PhanthomJay

Looks good .

3. Dec 25, 2013

### Speedking96

I just have a quick question. Suppose there was friction. And they asked for the total work done... would I have to subtract the work by friction from the total amount of Joules? Do we add "work" like vectors?

Thank you.

4. Dec 25, 2013

### haruspex

Work is a scalar.

5. Dec 25, 2013

### PhanthomJay

If there were friction, the total work done would be the algebraic sum of the work done by the train and the work done by friction .
Work , like energy, is a scalar quantity, not a vector quantity. It can have a positive or negative value, but it has no direction. In your example, yes, you would have to subtract the work done by friction from the work done by the engine to get the total, or net, work done on the train .

6. Dec 26, 2013

### Speedking96

For example, if the work done by friction was 100 000 Joules, then the total work would just be:

20 000 000 Joules + 100 000 Joules = 20 100 000 Joules.

However, if I wanted to find the total work done on the train, I would have to subtract the 100 000 Joules as you had said.

7. Dec 26, 2013

### haruspex

I find the question somewhat ambiguous. Does 'train' include the locomotive? These two excerpts make me think it's just the four cars:

8. Dec 27, 2013

### PhanthomJay

No!
Let's start from scratch with a different problem, hopefully clear of ambiguities. Suppose that a horizontal force F was applied to an object of mass M sitting on a horizontal friction-less surface.

If F = 20 000 N and M = 1 000 kg ,

a) How much work is done by the Force F on the object after it has accelerated over a distance of 1 km?
b) What is the total work done on the object?

For part a), you reason that the work done by F is F.d = 20 000 000 J.
For part b), the total work is the algebraic sum of the work done by all the forces acting on the object. The other forces acting on the object are the objects weight and the normal force of the floor on the object, neither of which does any work on the object (convince yourself of this). Thus, the total work is just the the work done by F, or 20 000 000 J, same answer as part a. I think you understand this already.

Now, same problem, same distance traveled, but lets add friction, assume a small 100 N friction force opposes the motion. The work done by F, the applied force, is still 20 000 000 J. The work done by friction is -100 000 J. Please note and understand the use of the minus sign here. The total work done on the object is thus 19 900 000 J (just add them up algebraically). No way no how can you ever say that the total work done on the object is 20 100 000 J.