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

In summary, a 900 kg car parked on a hill with a 15 degree angle and a coefficient of rolling friction of 0.1 begins to roll down the hill due to brake failure. Using the equations F=ma and v2=2aΔx, the car's speed after rolling 20m is calculated to be 8 m/s.
  • #1
Medgirl314
561
2

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?

Homework Equations


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!
 
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  • #2
Is the piano in the car ?
 
  • #3
Let's assume it's not for this situation. ;-) Is my answer correct?
 
  • #4
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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  • #5
Because I have all the information needed to solve. Why not?
 
  • #6
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?
 
  • #7
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! :smile:

Edit:Here'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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  • #8
Yes, yes. You shouldn't believe everything 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 ;-)
 
  • #9
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. :smile:

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=8Does that look better? This is why it's helpful to get an outside opinion. :smile:

Edit: LaTex failed ME. I see it now.
 
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  • #10
BvU said:
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?
 
  • #11
BvU said:
Yes, yes. You shouldn't believe everything 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.
 
  • #12
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...
 
  • #13
Yup, you got it! Perhaps I should have remembered to type out the full equation. :smile:

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.)
 
  • #14
Well done! On to the next. Car+piano and a pulley somewhere ;-)
 
  • #15
Thanks! Yeah, sure, I'll add it to my to-do list right after "Ride a unicorn on my own personal rainbow". ;)
 

FAQ: Physics Honors-Inclines-Car Rolling Down a Hill With Friction

1. How does friction affect the speed of a car rolling down a hill?

Friction is a force that opposes motion, so it will slow down the car as it rolls down the hill. The amount of friction depends on the surface of the hill and the weight of the car.

2. What is the relationship between the angle of the incline and the acceleration of the car?

The steeper the incline, the greater the acceleration of the car due to the force of gravity. This is because the component of gravity acting parallel to the incline increases with a steeper angle.

3. How does the mass of the car affect its motion down the incline?

The mass of the car affects the force of gravity acting on it, which in turn affects its acceleration down the incline. A heavier car will have a greater force of gravity and therefore a greater acceleration.

4. What is the difference between static and kinetic friction when it comes to a car rolling down an incline?

Static friction is the force that prevents an object from moving when a force is applied to it. In the case of a car rolling down an incline, static friction keeps the car from sliding down the hill until it reaches a certain angle. Once the angle is surpassed, the car will begin to move and kinetic friction comes into play, slowing down the car's motion.

5. How does the presence of friction affect the potential and kinetic energy of the car?

Friction converts some of the car's potential energy (due to its position on the hill) into kinetic energy (due to its motion). This means that the car will have less potential energy and more kinetic energy as it rolls down the hill. Friction also dissipates some of the car's energy as heat.

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