Conservation of Energy - Friction - Inclined Slope

AI Thread Summary
A 2.10 x 10^3 kg car starts from rest on a 4.7 m driveway inclined at 20°, facing an average friction force of 4.0 x 10^3 N. The discussion centers on calculating the car's speed at the bottom of the slope using energy conservation principles, where initial energy equals final energy plus work done by friction. The original poster struggles with their calculations, producing an unreasonably high final velocity of 588 m/s and an incorrect work value for friction. Contributors clarify that since the friction force is provided, there's no need to calculate it using the normal force or coefficient of friction. The focus remains on correctly applying the given values to find the car's speed.
dandy9
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Homework Statement


A 2.10 multiplied by 103 kg car starts from rest at the top of a 4.7 m long driveway that is sloped at 20° with the horizontal. If an average friction force of 4.0 multiplied by 103 N impedes the motion, find the speed of the car at the bottom of the driveway.

Homework Equations


Ei = Ef
W = Fcosxd

The Attempt at a Solution


I can't tell you how frustrated I am with this problem. I've done it four times already and can't seem to get the right answer.

My best attempt at solving it was as follows:
W(of friction force) = F(of friction) cosx d
Solving for W(of friction force) I got -363570833.8 --> a ridiculously big number that doesn't seem in the least bit to be correct.

Then I used initial Energy = final Energy + W(of friction force) and found the final velocity to be 588m/s. Which, again, is way too fast. My initial energy was potential and my final energy was kinetic.

If you could please help me out within the hour I would really appreciate it. I don't see where I'm going wrong.

Thank you!
 
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dandy9 said:
My best attempt at solving it was as follows:
W(of friction force) = F(of friction) cosx d
Solving for W(of friction force) I got -363570833.8 --> a ridiculously big number that doesn't seem in the least bit to be correct.
How did you get this answer? What numbers did you plug in?
 
My equation was:
N = normal force
u = coefficient of kinetic friction

W = uNcosxd
W = (4E3)(2.1E3)(9.8)(cos20)*(cos180)(4.7)
 
dandy9 said:
My equation was:
N = normal force
u = coefficient of kinetic friction

W = uNcosxd
W = (4E3)(2.1E3)(9.8)(cos20)*(cos180)(4.7)
The friction force is given as 4.0E3 and the distance is 4.7--that's all you need. (You don't have to calculate friction based on normal force.)
 
Hmm... Thanks.
But am I wrong in thinking that friction = uN
and because this is on an inclined slope you can't take the regular m*g, you have to take into account the angle - which is where I got my mg(cos20)...
Let me see how it goes.
 
dandy9 said:
But am I wrong in thinking that friction = uN
No, that's still true. But irrelevant here, since they give you the friction. (You aren't given any coefficient of friction.)
and because this is on an inclined slope you can't take the regular m*g, you have to take into account the angle - which is where I got my mg(cos20)...
That's all true, but in this problem you don't need that--they tell you the friction!
 
Yes! Thank you so much.
I really appreciate your time and your dedication to help students.
 
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