# Work and friction

1. Feb 27, 2005

### xelda

Whis is the minimum work needed to push a 1068 kg car 305 m up a 11.0 degree incline if the effective coefficient of friction is 0.26?

I have a hard time understanding why the normal force does not equal mg? How do I find the normal force? Or am I looking at this problem incorrectly?

2. Feb 28, 2005

### Galileo

Draw a picture of the car on the incline.
Gravity acts straight down vertically, but the normal force is perpendicular (normal) to the surface. Since the incline makes an angle with the vertical, the normal force acting on the car also makes an angle with the gravitational force.
Try to find the component of the gravitational force perpendicular to the surface (or parallel to the normal).

3. Feb 28, 2005

### dextercioby

Apply the theorem of variation of KE...It will be immediate.

Daniel.

P.S.Compute the forces correctly...Take Galileo's advice.He's really keen on inclines...:tongue2:

4. Feb 28, 2005

### ramollari

What do you mean by this?
It is just a matter of net force along the direction of motion and distance also.

5. Feb 28, 2005

### dextercioby

The minimum work done will be the work one has to do is simply compensate the work done by gravity & frition force...Sure,in this case,bacause,the initial & final states are not specified (namely the veloities being given),one cannot use the theorem...

So my advice was not lucrative for this problem,sorry.

Daniel.

6. Feb 28, 2005

### xelda

I would assume the normal force is mg x sin theta, but I was told this was incorrect?

7. Feb 28, 2005

### ramollari

No man. If $$\theta$$ is the angle that the incline makes with the horizontal then you take $$cos\theta$$! Draw the picture and see why.