# Homework Help: Minimum Work needed?

1. Jan 5, 2009

### Littlemin5

1. The problem statement, all variables and given/known data
what is the minimum wirk neede to push a 1000kg car 300m up a 17.5o incline?

Part a. Ignore friction

2. Relevant equations
So we are allowed to use the general equation

W=FdCos $$\theta$$

3. The attempt at a solution

So I thought you would just do:

W= (1000)(9.8)(300)(cos17.5)
W=2803927Joules

However according to my teacher we should be getting
W= 8.8 x105 Joules

Does anyone know what I am doing wrong?

2. Jan 5, 2009

### Dick

Draw a picture. Are you sure cos is the right trig function to be using? The force is aligned with the direction of the car's motion so W=F*d. How does the angle of the slope affect F? Split the force into components.

3. Jan 5, 2009

### Littlemin5

Oh, so you would have to use sin instead of cos. That makes more sense. I didnt know that you could interchange the trig function in the equation.

Thanks so much for your help!

4. Jan 5, 2009

### Dick

You can't interchange the trig functions in the equation!! What happens when you are pushing the car up the incline is that the only force you have to overcome is the component of m*g that is tangent to the road. That's F_g=m*g*sin(incline angle). In the W=F*d*cos(theta) the theta is 0, since we are pushing in the same direction the car is moving. Those are two DIFFERENT angles. You aren't just substituting 'sin' for 'cos'. Try to be clear on this.

5. Jan 5, 2009

### Littlemin5

But I do have one more question if I was told that in the next part there was an effective coefficent of friction of .25 , wuld I just multiple my answer for Part 1 by .25?

6. Jan 5, 2009

### Dick

No. Not at all. Now you have to figure out the normal force before you can compute the frictional force. Can you do that?

7. Jan 5, 2009

### Littlemin5

Wouldn't you do 9800Cos17.5=9346.26N

so your answer there would be the Normal Force?

8. Jan 5, 2009

### Dick

Right. So now get the frictional force. The total force you have to push up the hill then the tangential force (as in the first problem) PLUS the frictional force.

9. Jan 5, 2009

### Littlemin5

so frictional force is 2336.565 and then I add that to 9800 which equals

12136.565.

From there I would do
(12136.565)(300)(sin17.5)

Right? And the answer I get would be my answer?

10. Jan 5, 2009

### Dick

No again. F_total=F_tangential+F_friction. Ok, F_friction is 2336N. The gravitational force component you are opposing is m*g*sin(17.5). Now take F_total*d

11. Jan 5, 2009

### Littlemin5

Wait I don't really understand the last comment you made. Could you please explain it in a bit more detail?

12. Jan 5, 2009

### razored

Here is another perspective. The displacement occurs at 17.5 degrees from the horizontal. The force, that is gravitational force, is acting down. We ignore the normal work because it has no component of force in the displacement (cos90=0). So, since mg is acting down, and the displacement is at 17.5 above horiztonal, we use the FDcosO. We know the force, mg; we know the displacement, 300m, and we know the angle between them, 17.5 + 90. So.. it should be 1000kg * 300m * 9.8 * cos (17.5 + 90). I think this is right. This is assuming no friction. Also, the work is negative but magnitude is positive.

13. Jan 5, 2009

### Dick

You could certainly do it that way. But it's pretty usual with inclined plane problems to split the m*g force into tangential and normal components. Going up the plane you only do work against the tangential component of m*g and the friction force. Add them and multiply by the distance.

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