Finding distance using Work

dks2114 · #1 Oct 8, 2007, 04:32 PM

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?

ebaines · Oct 9, 2007, 07:49 AM

I think youre 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.

dks2114 · Oct 9, 2007, 09:12 AM

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

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

ebaines · Oct 9, 2007, 10:57 AM

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.

dks2114 · Oct 9, 2007, 11:17 AM

Ohhhh thank you VERY much!

ebaines · Oct 9, 2007, 12:21 PM

You're welcome!



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