Power used to move a car up an incline with friction

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Homework Statement


a 1500 kg car is moving up a 25 degree incline at a constant speed of 30 mph. the tires and road have an effective friction coefficient of u = 0.25 . determine the power output the engine must provide to keep the car moving at this speed. express your final answer in units of watts and horse-power. if an answer cannot be found then briefly explain why this is so.

Homework Equations


Wnet = E_final - E_initial
Power = Work / Time
Power = Force (dot) Velocity
KineticE= 1/2 * Mass * Velocity^2
F_friction= u * F_normal

The Attempt at a Solution


I assumed the answer can be found...

step 1:
I initially drew a Free Body Diagram:
Normal - perpendicular to the incline pointing above it
Push - parallel to the incline pointing up it
Friction - parallel to the incline pointing down it
MG - straight down

Step 2:
This is where I am stuck. What I believe I am trying to find is the energy consumed per time unit overcoming friction PLUS the energy consumed per time unit to keep the car moving at 30mph. If this is correct I still am struggling with how to use the equations to find power.(mostly how to go from Force to Work to Power)

my attempt to set up the equation:

30mph = 13.4112m/s

F_f = .25 * sin(65) * 1500 * 9.8 = 3330.68 N
W_f = 3330.68 * delta X
Above where would I even get (delta X)

KE= (1/2) * 1500 * 13.4112^2 = 134895 J

W_f + KE = Total Energy

after this I don't have a time defined to turn work to power
 

Answers and Replies

  • #2
Orodruin
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Power is energy per unit time. How far does the car move in one unit time?

Are you also not forgetting one force that is performing negative work? (And thus must be counteracted by the engine)
 
  • #3
BvU
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Hello Dan3 and welcome to PF.

Better to keep working in expressions in terms of the variables. That way you can check dimensions and things that cancel don't unnecessarily fuzz the results.

Now for this car: Wnet = Efinal - Einitial is energy and you want power = energy/time.
So if you can establish the time between your initial and final situations, you're in business.

Then: I miss potential energy in your balance !

An energy balance is one way to fix this one. Another (in fact the same) way would be to use your Power = Force (dot) Velocity
 
  • #4
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Power consumed ( Watts ) = force overcome ( N ) * velocity ( m/s ).
At constant speed (non-accelerating) means the for and against forces are equal.

Find the total force against ( rolling resistance plus gravitational )
( this = the force to be overcome by the drive force )
Rolling resistance = m * g * Cosine 25 ° * µ
Gravitational = m * g * Sine 25 °

Watts / 745.7 = hp
 
  • #5
haruspex
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Find the total force against ( rolling resistance plus gravitational )
Sure, but we have no information on rolling resistance, so maybe take it as zero. It is nothing to do with coefficient of friction. The tyres roll on the road, so no work is done against friction there.
 
  • #6
Orodruin
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Sure, but we have no information on rolling resistance, so maybe take it as zero. It is nothing to do with coefficient of friction. The tyres roll on the road, so no work is done against friction there.

My guess would be that the problem author was sloppy (or ignorant) and that rolling resistance is what (s)he means by "effective friction".
 
  • #7
haruspex
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My guess would be that the problem author was sloppy (or ignorant) and that rolling resistance is what (s)he means by "effective friction".
Perhaps, but 0.25 is a very high value for that. It is equally believable that it is a test of the student's understanding, and that treating it as resistance will lose marks. I know what you mean about 'effective', though.
The important thing here is that DanDanDan is not left believing that it is right to treat friction as a force opposing motion in rolling contact.
 
  • #8
Orodruin
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Perhaps, but 0.25 is a very high value for that. It is equally believable that it is a test of the student's understanding, and that treating it as resistance will lose marks. I know what you mean about 'effective', though.
The important thing here is that DanDanDan is not left believing that it is right to treat friction as a force opposing motion in rolling contact.

I fully agree. The sad part is that if it is the other way around he will be marked wrong for not taking it into account, leaving a student perplexed and not knowing in or out. I also find it completely plausible that someone who does not differentiate friction from rolling resistance should pick a too high value for it. But back to the subject at hand.
 

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