# Please check my work Friction problem

## Homework Statement

a car's engine, producing a constant power, propels the car up a hill at a constant speed and down the hill at double the speed. if the coefficient of kinetic friction between the car and the road is 0.28, what is the angle of incline of the hill. assume the hill is symmetric.

## The Attempt at a Solution

While going upwards, friction and weight component are acting downwards.
∴Ef=Ff+mgsinθ
Ef=μN+mgsinθ
=μmgcosθ +mgsinθ

While going downwards, the frictional force reverses direction. but the velocity is double....So i am assuming that the downwards force is double.
∴2Ff=mgsinθ+Ef
Ef=2μmgcosθ-mgsinθ

equate the two expressions:
μmgcosθ +mgsinθ=2μmgcosθ-mgsinθ
2mgsinθ=2μmgcosθ
tan θ=μ=0.28
θ=tan(-1) 0.28
=15.64 degrees

PhanthomJay
Homework Helper
Gold Member
I don't understand how kinetic friction comes into play if the tires are not skidding; the resistance must be provided by rolling or axle friction, and air drag, but I'll assume that the use of the word 'kinetic' is correct. I'm not following your equations or logic, why do you say the downward force (presumably from the cars engine) is doubled? If the power it delivers is constant, but its speed down is doubled, then what is the relationship between the force it delivers up the incline vs. down the incline?

I don't understand how kinetic friction comes into play if the tires are not skidding; the resistance must be provided by rolling or axle friction, and air drag, but I'll assume that the use of the word 'kinetic' is correct. I'm not following your equations or logic, why do you say the downward force (presumably from the cars engine) is doubled? If the power it delivers is constant, but its speed down is doubled, then what is the relationship between the force it delivers up the incline vs. down the incline?

that might be the case but axle friction and air drag have not been covered. to be honest, my solution was a stab in the dark...

PhanthomJay
Homework Helper
Gold Member
that might be the case but axle friction and air drag have not been covered. to be honest, my solution was a stab in the dark...
Well it wasn't too bad of a stab considering the ambiguity in the problem statement. You might want to try again, noting that power = Fv, which is constant up or down the incline, and use Newton 1 to sum all forces equal to zero. Watch your math and plus and minus signs.

Well it wasn't too bad of a stab considering the ambiguity in the problem statement. You might want to try again, noting that power = Fv, which is constant up or down the incline, and use Newton 1 to sum all forces equal to zero. Watch your math and plus and minus signs.

P1/P2=F1v1/F2v2
∴F2*2v1=F1v1
F2=0.5F1
so the force while going down is half the force while coming up.
Ff=mgsinθ+0.5Ef
Ef=2Ff-2mgsinθ
equating the equations:
2μmgcosθ-2mgsinθ=μmgcosθ +mgsinθ
μmgcosθ=3mgsinθ
tanθ=μ/3
θ=tan-1(0.0933)
=5.33 degrees..

PhanthomJay