# Physics Honors-Inclines-Car Rolling Down a Hill With Friction

## Homework Statement

A 900 kg car is parked on a hill when its brakes fail. It begins to roll down the hill,which is at a 15 degree angle above the horizontal.The coefficient of rolling friction is 0.1. What is the car's speed after it has rolled 20m?

F=ma
v2=2aΔx

## The Attempt at a Solution

a=9.8(sin15)-0.1(9.8)(cos15)
v2=2.54-0.947
v=1.6 m/s

Looks good to me! Is it right?

Thanks!

BvU
Homework Helper
Is the piano in the car ?

Let's assume it's not for this situation. ;-) Is my answer correct?

BvU
Homework Helper
Why v2=2aΔx ?
With a=9.8(sin15)-0.1(9.8)(cos15) you find a = 2.54-0.947 = 1.59 m/s2 along the slope downwards.

Why do you continue with v2 = 1.59 m/s ? It's not the same thing and it's also not in agreement with the (erroneous) second equation. So why ?

Stick to ##a = mg \sin\theta - \mu \cos\theta## !
Now you have a constant acceleration along the slope. initial speed 0.

You know from past exercises an expression for v(t) and for s(t). If you forgot, learn them by heart and never ever forget them again.
s(t) = 20 gives you t, v(t) is then the desired answer. Off you go !

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Because I have all the information needed to solve. Why not?

BvU
Homework Helper
I've never come across v^2 = 2aΔx, so I am curious what it means and where it comes from. Would you have a derivation?

Ah! Okay. The full equation is actually v2=v02+2aΔx, which is an equation for final velocity. I dropped the v02 because it's zero in this situation. There are so many different variables for the same term, so in case this still looks unfamiliar:

v= Final velocity.
v0= Initial velocity.
a=Acceleration.
Δx=Change in distance.

Hope that helps! Edit ere's a link to my equation sheet. These are just common, general equations. I found the one I used here.
http://derekowens.com/access2/physics/EquationSheet.pdf

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BvU
Homework Helper
Yes, yes. You shouldn't believe everthing these physicists say (type).
The ##{1\over 2} m v^2 - {1\over 2} m v_0^2 = mg \Delta h## is an energy balance. On the left you have the change in kinetic energy, on the right you have the change in potential energy.

If you understand what you are doing, no problemo. But in the relevant equation there is a ##\Delta x## that mysteriously disappears. What about thay guy ?

The eqn sheet is nice, if you know what things refer to. Would be even nicer if the eqns were numbered, then we wouldn't have to retype them all the time ;-)

Well, the physicist that typed the equations was my renowned physics teacher, so I think it's okay to use the equations he likes for his class. Is LaTex failing you? The equation is a bit difficult to read.

Ah, I see what you mean. That part was user error. ;-)
v^2=2aΔx
v^2=2(1.6)(20)
v^2=64
v=8

Does that look better? This is why it's helpful to get an outside opinion. Edit: LaTex failed ME. I see it now.

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Why v2=2aΔx ?
With a=9.8(sin15)-0.1(9.8)(cos15) you find a = 2.54-0.947 = 1.59 m/s2 along the slope downwards.

Why do you continue with v2 = 1.59 m/s ? It's not the same thing and it's also not in agreement with the (erroneous) second equation. So why ?

Stick to ##a = mg \sin\theta - \mu \cos\theta## !
Now you have a constant acceleration along the slope. initial speed 0.

You know from past exercises an expression for v(t) and for s(t). If you forgot, learn them by heart and never ever forget them again.
s(t) = 20 gives you t, v(t) is then the desired answer. Off you go !

This thread is becoming hard to follow, with all the edits made. I'm not sure when you edited this post, but if you check the post right before this, we may have come to a conclusion. Could you please explain why the equation for velocity is wrong? Do we need to involve mass?

Yes, yes. You shouldn't believe everthing these physicists say (type).
The ##{1\over 2} m v^2 - {1\over 2} m v_0^2 = mg \Delta h## is an energy balance. On the left you have the change in kinetic energy, on the right you have the change in potential energy.

If you understand what you are doing, no problemo. But in the relevant equation there is a ##\Delta x## that mysteriously disappears. What about thay guy ?

The eqn sheet is nice, if you know what things refer to. Would be even nicer if the eqns were numbered, then we wouldn't have to retype them all the time ;-)

Another edit. XD Haha, I didn't write it, but retyping them seems easier than converting the document and then having to go find it to copy and paste all the time.

BvU
Homework Helper
Oh bugger, I'm old and have to type with two fingers. Can never win a speed typing contest.
With ##v=v_0+at##, ##v_0=0## so ##v=at## so ##v^2=a\ at^2## and ##x=x_0+v_0 t+{1\over2} at^2## and ##x_0=0## so ##x = \Delta x## and ##v_0=0## so ##x={1\over2} at^2## I can understand ##v^2=2a \Delta x##.

And yes, 8 m/s (7.97 is wat I see here).

You're getting good at this !

Put ## \mu=0## and find 10 (or 10.07) m/s Oops, against PF rules....

Yup, you got it! Perhaps I should have remembered to type out the full equation. Haha! That's okay, I have plenty of other problems to work on while I wait.

Thanks, I probably rounded a bit.

a=9.8(sin15)
a=2.54
v^2=2aΔx
v^2=2(2.54)(20)
v^2=101.6
v=10.08 m/s (Rounding a bit.)

BvU