How Far Will a Car Slide Uphill vs Downhill When Braking?

  • Thread starter Thread starter domagoj412
  • Start date Start date
  • Tags Tags
    Car Hill
AI Thread Summary
A car traveling uphill at 60 km/h on a 17° incline will slide approximately 16.34 meters after braking, while the same car traveling downhill will slide about 50.31 meters. The calculations involve converting speed from km/h to m/s and applying energy conservation principles. The equations used include kinetic energy, potential energy, and frictional force. The discussion confirms the correctness of the calculations after adjustments were made. Accurate results depend on proper unit conversion and application of physics principles.
domagoj412
Messages
9
Reaction score
0

Homework Statement


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...)

Homework Equations


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


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
 
Physics news on Phys.org
domagoj412 said:

Homework Statement


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...)

Homework Equations


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


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.
 
Yes, I got the different result:

16.34 m for uphill and 50.31 m for downhill.
 
domagoj412 said:
Yes, I got the different result:

16.34 m for uphill and 50.31 m for downhill.

Your answer is correct now.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
1
Views
2K
Replies
14
Views
2K
Replies
29
Views
3K
Replies
7
Views
3K
Replies
8
Views
2K
Replies
8
Views
5K
Replies
8
Views
8K
Back
Top