End position of braking car on an icy incline

[imatreyu]

Homework Statement

A car with a speed of 40.0 km/h approaches the bottom of an icy hill. The hill has an angle of inclination of 10.5 degrees. The driver applies the brakes, which makes the car skid up the hill. If the coefficient of kinetic friction between the ice and the tires if 0.153, how high, measured along the incline, is the car on the hill when it comes to rest?

Homework Equations

F=ma. . .and. . .eventually kinematics?

The Attempt at a Solution

Wy= Wsin(theta)--> mgsin(theta)
Wx= Wcos(theta)--> mgcos(theta)

Y: N-mgsin(theta) = 0 (there is no y acceleration?)
N= mg sin(theta)

X: N(mu) - mgcos(theta) = ma
mgsin(theta)mu-mgcos(theta)=ma
--> m cancels out-->
gsin(theta)mu-gcos(theta)=a


Where can I go from here? What I have I done wrong? :( Thank you in advance!

 

[rl.bhat]

Since car is going up, the friction force and the component of weight along the inclined plane act in the same direction.

 

[imatreyu]

But doesn't the friction force make the car skid up the hill (which would make it opposite from the weight x component)?

 

[rl.bhat]

But doesn't the friction force make the car skid up the hill (which would make it opposite from the weight x component)?

The car moves up due to its initial KE. Frictional force acts in opposite direction to the motion of the car. The car is moving up. So Fr acts in the downward direction and brings the car to rest.

 

[imatreyu]

So, once I correct the direction regarding friction, should the rest of the process lead me to the solution?

(Supposedly 19 m)

Because I did correct the direction of the friction, and the acceleration I am getting is suspiciously near 9.8 m/s2. . I think there must be a mistake elsewhere. .?

 

[rl.bhat]

I am getting correct answer. Show your calculations.

 

[imatreyu]

Y: N-mgsin(theta) = 0 (there is no y acceleration?)
N= mg sin(theta)
---->
X: N(mu) + mgcos(theta) = ma
mgsin(theta)mu+mgcos(theta)=ma
--> m cancels out-->
gsin(theta)mu+gcos(theta)=a

so:

9.8sin(10.5)(.153) + 9.8cos(10.5) = a
a= 9.90914. . .m/s^2

vi= 40.0 km/h --->11.1111m/s
vf= 0 m/s

vf^2 = vi^2 + 2ad
0^2= (11.111) + 2 (9.90914) d
d= - 0.56064:( Something is wrong. . .

Still getting wrong answer.

 

[rl.bhat]

N(mu) + mgcos(theta) = ma

Normal reaction is mg*cosθ

And vf^2 = vi^2 - 2ad

 

[housemartin]

i think in you got x components of forces wrong. Component of "gravity (mg)" along the hill is not mgcos(teta). Check your diagram. And one more important thing, before writing Newtons second law, always indicate your coordinate axis ;] this way you can keep track of positive and negative acceleration etc.

 

[imatreyu]

rl.bhat,

mu (-9.8) cos10.5 - 9.8sin10.5 =a
a= 03.08675439

--> 0= 11.111^2 + 2 a d

solve for d= 19.99. . .

I think this is about right?

housemartin, The coordinate axis is aligned so that the positive x is going up the direction of the incline. Is the component of gravity parallel to the hill actually supposed to be mgsin(theta)?
Thank you to all. . .

 

[housemartin]

so if positive x is uphill component of gravity points in which direction? +x or -x? And in which direction does friction force points?

 

[imatreyu]

Gravity is pointing negative. Friction forces the car positive.. . .I don't even understand how "The driver applies the brakes, which makes the car skid up the hill. " is possible. .

 

[housemartin]

Friction force always opposes motion - it makes things go slower - velocity decrease, and that mean what? it's not in the positive direction ;]
If driver didn't applied brakes, car would slow down only because of gravity, which component uphill is negative. With brakes, another force "helps" gravity to slow the car faster.

 

[imatreyu]

Aha! Thank you, rl.bhat and housemartin. =)