Car Braking (On Inclined Slope)

AI Thread Summary
A 900kg car moving up a 15-degree incline at 40 m/s skids to a halt over 40 meters, prompting calculations of total work done. The discussion revolves around using kinetic energy and gravitational potential energy to determine work, with confusion about whether to include the vertical component of gravity. Participants clarify that the work done against friction is distinct from the work done by friction, emphasizing that friction always opposes motion. The calculations yield a work done against friction of 628.7 kJ, leading to a frictional force of 15.7 kN, although concerns arise about the high coefficient of friction calculated. The problem's wording is criticized for lack of clarity regarding the forces involved.
ariarch
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Homework Statement



A 900kg car is moving up a 15degree inclined slope at 40ms-1. The driver slams on the brakes, skidding to a halt 40m along the road.

Calculate the total work done by the car.

Homework Equations



This is what I am not sure of. As I am not told whether or not the acceleration is constant, I doubt I can use f=ma.

I'm therefore thinking that I can use:

K.E + GPE?

So, W.D = 1/2(mv12-mv22) + mgh

where m = 40sin15

The Attempt at a Solution



I therefore solve, getting total work done as:

1/2(900*402) + 900*9.81*40sin15.

Or do I have to use the vertical component of gravity? I am perplexed, fine mechanics the most difficult part of Physics, and this is driving me mad!

Thank you.
 
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Welcome to PF, Ariarch!
It is a confusing little problem to me. It could be
work done by car = energy converted to heat + potential energy gained
or just
work done by car = energy converted to heat
I don't think the wording makes it clear which. In the first case the calc is just work done = ½⋅mv². In the other it is ½⋅mv² - mgΔh
My guess would be the latter since the first case is too easy.
 
Thank you:)

I am still a little confused. Why would it be -mgh? as I thought work would needed to have been done to elevate the car, therefore meaning it would be +mgh??
 
Oh yes, and I don't know if this would make it clearer but I also have to work out the work done by the weight and friction..
 
OK, well I understand that the total work done on the car is now equal to the change in kinetic energy, which must equal 1/2(mv^2) = 720kJ.

If I divide this by the distance traveled (40m), I can find that the overall force acting on the car is equal to 180kN.

However, if I calculate the components of weight and friction opposing the direction of motion, this does not add up! :S :S
 
work done against friction = energy converted to heat
= original KE - remaining PE
= ½⋅mv² - mgΔh
 
Of course, that makes a lot of sense. So that will be the total work done on the car? and from that I can calculate the Work done from the separate components (weight and friction)?
 
I can't say it is the total work - remember my confusion mentioned in my first post. It is certainly the work done against friction.

The force of friction is a function of the normal force and the coefficient of friction but I don't think you need to get into those details in this question. It is quite complicated because of the angle of the hill - the normal force is not equal to the weight.
 
Hmm yes, I haven't been able to encounter a problem similar to this anyway (it is all braking on a flat plane, not on an incline).

But I am probably being a little slow here, work done against friction is not the same as work done by friction?

and is it true to say that while the car is braking, it is the kinetic friction that is doing the actual work (as well as the x-component of weight)?
 
  • #10
But I am probably being a little slow here, work done against friction is not the same as work done by friction?
Friction doesn't do work - it can't make a car go faster. It always takes energy away so it makes more sense to say "work done against the force of friction".

Oops, I wasn't thinking very deeply about the friction! I suppose most of it is friction in the brake system rather than friction with the road or air. The road friction is static because the wheels are not slipping on the road. The disk brake is kinetic, as is the wind.
 
  • #11
Ah ok then, so the question is indeed badly worded... I'll be having strong words with my lecturer!

Ok, I think I am supposed to neglect air resistance, so there should only be Kinetic Friction in the wheels right?
 
  • #12
With the information given, there is no way to sort out how much of the friction is due to air resistance. You can get the overall friction work and the total retarding force, but not the part that is due to the air or road and you can't find the coefficient of friction.
 
  • #13
Ok, I've been thinking about this allll night (didn't get much sleep!)

I've come to a rather unsound conclusion... While more work done may be required to move the car up the hill (in the form of mgh), the action of gravity opposing the direction of motion will mean that LESS work will need to be done in order to slow the car down, therefore they will cancel out.

So, W.D = dK.E = 0.5*900*402

Then, if I resolve components of weight and friction along the x-direction, I see that Work Done by the weight of the car = mgsin15 * 40 = 91.3kJ.

Therefore Work Done by friction must be W.D(total)-W.D(weight) = 628.7kJ. From this, I can use the equation W.D = F*d to find the magnitude of the frictional force (15.7kN). The only thing is, this would mean that coefficient of static friction is 1.8 (which seems a little high to me).

But hey ho, does this seem correct?
 
  • #14
Good morning!
I see that "work done by the weight of the car" means the mgh potential energy gained; didn't understand what you meant by that phrase before. I agree with that 91.3 kJ and with the 628.7 kJ for the work against friction.

I just read the question again and saw the "skidding to a halt in 40 m". That means the brakes locked the wheels and we have kinetic friction and all the energy was lost slipping on the road, none on the brakes because the wheels weren't turning. I was confused about that yesterday! So we have the Ff = 15.7 kN but what is the normal force pressing the car against the road? Less than mg because only a component of mg is perpendicular to the road, mg*cos(15) I think. That makes the coefficient of kinetic friction a wee bit higher than 1.80. It does seem high.
 
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