Question regarding distance traveled up an incline?

  • Thread starter choyphin
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  • #1
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Homework Statement


After a bobsled race, the sled and riders have been partially slowed down up in an icy incline, but they need to be brought to a stop. This happens on a portion of track inclined at 30 degrees that has a coefficient of kinetic friction of 0.60. The sled enters the incline at 25.0 m/s. How far does it travel along the incline before stopping?


Homework Equations


(1/2)mv(final)^2 - (1/2)mv(initial)^2 = Ffriction


The Attempt at a Solution


Im not sure how to incorporate the angle of the incline, but as an attempt, i did
1/2vi^2=FG(cos 30) d
and i got 88.3 m. I'm not sure if that's right...
 

Answers and Replies

  • #2
Doc Al
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Looks like you're using an energy approach. Good. But don't forget gravitational PE, which increases as the sled goes up the incline.
 
  • #3
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How would you incorporate gravitational PE? and also, would the formula
d = vi^2 / [2g(sin30 + ucos30)] work?
 
  • #4
Doc Al
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How would you incorporate gravitational PE?
What's the definition of gravitional PE? How do you calculate it?
and also, would the formula
d = vi^2 / [2g(sin30 + ucos30)] work?
I would advise against looking for a formula to plug into unless you fully understand what the formula means. Better to stick with the basics and derive your own formula, tailor-made to the specific problem. (That formula is close, but not quite right, anyway.)

The basic idea that you want to use is energy conservation:
Energy(initial) + Work done by friction(which is negative) = Energy(final)

Here "Energy" means total mechanical energy: KE + PE.
 

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