Calc Work Needed to Push Car up 11.0 deg Incline

[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?

 

[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).

 

[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...

 

[ramollari]

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

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

 

[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.

 

[xelda]

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

 

[ramollari]

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

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.