• Support PF! Buy your school textbooks, materials and every day products Here!

W=FD with a µ Curveball

  • #1
I have honestly been trying to figure this problem out for over an hour now, and every time I try to solve for KE, I get a ridiculously high number.

Homework Statement


A race car having a mass of 1000 kg was traveling at high speed on a wet concrete road under foggy conditions. The tires on the vehicle later were measured to have µ = 0.55 on that road surface. Before colliding with the guardrail, the driver locked the brakes and skidded 100 m, leaving visible marks on the road. The driver claimed not to have been exceeding 65 miles per hour (29 m/s). Use the equation W=KE to estimate the driver's speed upon hitting the brakes.


Homework Equations


W=KE
KE=(mv^2)/2


The Attempt at a Solution


Every time I try to substitute and solve for KE, I'm getting a workload of over 300,000 N for the brakes, obviously impossible.
 
Last edited:

Answers and Replies

  • #2
diazona
Homework Helper
2,175
6
First of all, why is it "obviously impossible"? Secondly, how exactly are you getting that number? (i.e. show your work)
 
  • #3
Doc Al
Mentor
44,882
1,129
Every time I try to substitute and solve for KE, I'm getting a workload of over 300,000 N for the brakes, obviously impossible.
Show exactly what you are doing.

One formula you didn't list is the one you need to compute the work done by friction. What would that be?
 
  • #4
Ok,
KE=(.5)(mv^2)
KE=(.5)(1000 kg * (29 m/s)^2)
KE=420,500 N
W=KE
How can I possible have 420,500 N of work done by brakes?
 
  • #5
Or am I doing this wrong...
Since the force that stopped the car was frictional; won't I have to calculate the frictional force using the Newton weight of the car? And use the frictional force in the KE and Work formulas?
 
  • #6
53
0
Work is not just equal to K.E.

The work-energy theorem says:

W = delta K.E. = [1/2 mv(final)^2] - [1/2 mv(initial)^2]
 
  • #7
Doc Al
Mentor
44,882
1,129
Ok,
KE=(.5)(mv^2)
KE=(.5)(1000 kg * (29 m/s)^2)
KE=420,500 N
W=KE
How can I possible have 420,500 N of work done by brakes?
The unit of work/energy is Joules, not Newtons. And why do you think that that is an unreasonable amount of energy, or that brakes can't handle it? (Note that you don't need to calculate this energy to solve the problem.)

Or am I doing this wrong...
Since the force that stopped the car was frictional; won't I have to calculate the frictional force using the Newton weight of the car? And use the frictional force in the KE and Work formulas?
Absolutely. What does the friction force equal? (Symbolically, not numerically.) Set the work done by friction equal to the initial KE, then solve for the speed.
 

Related Threads for: W=FD with a µ Curveball

  • Last Post
Replies
2
Views
755
  • Last Post
Replies
2
Views
1K
Replies
0
Views
2K
Replies
6
Views
12K
Replies
1
Views
1K
  • Last Post
Replies
1
Views
16K
Replies
5
Views
574
  • Last Post
Replies
7
Views
11K
Top