Suppose a gravel ramp slopes upward at 6.0 degrees and the coefficient of friction is 0.40. Use work and energy to find the length of a ramp that will stop a 15000kg truck that enters the ramp at 35 m/s.


This is what I did:
I calculated Ff and got 6146.4N. Plugged that into the work formula. Set that equal to KE ((1/2)mv^2 = Fd)and solved for d (and got 1495m). Is this correct?

I think you're Ff is off - how did you calculate that? Should be Ff = .4 * 15,000kg * 9.8m/s^2 * cos(6 degrees) = 58477 N. Also, don't forget to include the work the truck does against gravity as it rolls up the ramp.

Why is it cos(6.0) and not sin(6.0) to find Fn?

4802

I thought the vector diagram looks like this. Did I put the angle in the wrong place or something?

The angle you marked in the figure is not 6 degrees: what you marked is the complimentary angle (i.e. 90 - 6 = 84 degrees). Sin(84) and cos(6) are the same.

Ohhhh thank you VERY much!

You're welcome!