Car driving up the hill...

[domagoj412]

1. The problem statement, all variables and given/known data
Car is driving up the hill with speed of 60 km/h. Angle of incline is 17°. Seeing barrier on the road, the driver suddenly stops (breaks the wheels) and starts to slide. If koef. of fiction is \mu=0.6 what is the distance between breaking point and the point when cars stops. Solve the problem with the same parameters when the car is driving down the hill.

(I hope you understand the question, it is a little difficult for me to translate it...)

2. Relevant equations
Again, I solve it but I'm not sure if is correct..


3. The attempt at a solution

Uphill:
Ek = Ep + Fk*s
1/2 mv^2 = mgh + \mu*m*g*cos(\alpha)*s
(h = s * sin(\alpha))
...
s = 32,7 m

Downhill:
Ek + Ep = Fk*s

[vij]

1. The problem statement, all variables and given/known data
Car is driving up the hill with speed of 60 km/h. Angle of incline is 17°. Seeing barrier on the road, the driver suddenly stops (breaks the wheels) and starts to slide. If koef. of fiction is \mu=0.6 what is the distance between breaking point and the point when cars stops. Solve the problem with the same parameters when the car is driving down the hill.

(I hope you understand the question, it is a little difficult for me to translate it...)

2. Relevant equations
Again, I solve it but I'm not sure if is correct..


3. The attempt at a solution

Uphill:
Ek = Ep + Fk*s
1/2 mv^2 = mgh + \mu*m*g*cos(\alpha)*s
(h = s * sin(\alpha))
...
s = 32,7 m

Downhill:
Ek + Ep = Fk*s

Your equations are correct. But the calculation appears to be incorrect. Probably you have not converted km/h into m/s. Try once again and finish both calculations.

[domagoj412]

Yes, I got the different result:

16.34 m for uphill and 50.31 m for downhill.

[vij]

Yes, I got the different result:

16.34 m for uphill and 50.31 m for downhill.

Your answer is correct now.