Stopping Distance: Car F, G, and H

  • Thread starter Thread starter Soaring Crane
  • Start date Start date
  • Tags Tags
    Stopping distance
Click For Summary
Cars F, G, and H, despite having different masses, will travel the same distance to skid to a stop when they slam on the brakes, assuming they have identical tires and thus the same coefficient of friction. The force of friction acting on each car is proportional to its weight, which balances out with their kinetic energy during braking. The equations provided demonstrate that the energy dissipated by friction is equal for all three cars, leading to the same stopping distance. The stopping distance can be calculated using the relationship between energy and force, specifically E/F. Therefore, all cars will stop over the same distance regardless of their mass.
Soaring Crane
Messages
461
Reaction score
0
Three cars (cars F, G, and H) are moving with the same velocity, and slam on the brakes. The most massive car is F, and the least massive is H. Assuming all 3 cars have identical tires, which car travels the longest distance to skid to a stop?

Will they all travel the same distance in stopping?
If mgh = mv^2/2 for each car:

F – (3mv^2)/2 = 3*mgh h = v^2/2g
G – (2mv^2)/2 = 2mgh h = v^2/2g
H – mv^2 = mgh h = v^2/2g

Thanks.
 
Physics news on Phys.org
The same tires implies the same coefficient of friction, u or \mu.

The force of friction applies, Ffriction = \mumg, and the energy dissipated is Efriction = Ffriction*d, where d is the distance traveled.

Find dF = dG = dH.
 
How do I find the energy of friction to find each car's d?

To find d, it would be E/F?

Thanks again.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
4
Views
1K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
5K
Replies
40
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K