# Work and energy problem

[SOLVED] work and energy problem

## Homework Statement

A car (m = 890.0 kg) travelling on a level road at 27.0 m/s (60.5 mph) can stop, locking its wheels, in a distance of 60.0 m (196.9 ft). Find the size of the horizontal force which the car applies on the road while stopping on the road. First solve this problem using work/energy concepts and then check your answer using kinematics/force law concepts.

I solved this by finding the acceleration and plugging into formula F=ma=5402N

Find the stopping distance of that same car when it is traveling up a 18.9deg slope, and it locks its wheels while traveling at 27.0 m/s (60.5 mph). Assume that muk does not depend on the speed.

## Homework Equations

3. The Attempt at a Solution [/b

I drew a free body diagram and came up with the following relationships
F-fk-mgsin(18.9)=ma where F=5402N
N-mgcos(18.9)=0

How do I solve for the horizontal distance? Is it just the vertical component of the Force?

Doc Al
Mentor
I drew a free body diagram and came up with the following relationships
F-fk-mgsin(18.9)=ma where F=5402N
That F=5402N doesn't belong here--that was the friction force from the first part.
Redo this without that extra force.
N-mgcos(18.9)=0
Good.
How do I solve for the horizontal distance? Is it just the vertical component of the Force?
Solve for the acceleration. (Or the net force parallel to the road.)

OK, so i have F-ukmgcos(18.9)-mgsin(18.9)=ma. Can I solve for uk from the previous part of the problem and use it in this equation? If so, I still have one equation and 2 unknowns (F and a).

To solve for x i think I can use the equation Vf^2=V0^2+2ax, where Vf=0 and V0 is given.

Doc Al
Mentor
OK, so i have F-ukmgcos(18.9)-mgsin(18.9)=ma.
What does the F stand for? You already have friction and weight accounted for.
Can I solve for uk from the previous part of the problem and use it in this equation?
Absolutely.
If so, I still have one equation and 2 unknowns (F and a).
Lose the F!

F is the horizontal force (same as what the first part of problem asked for)

Well, I tried losing the F, which doesn't make much sense to me, and have -ukmgcos(18.9)-mgsin(18.9)=ma --> a=-14.57 m/s^2

use Vf^2=V0^2+2ax --> (27)^2=-2ax --> x=25.01m which STILL isn't right! :(

and i found uk with this equation from the first part

5402-ukmg=ma where a=-6.075
uk=1.23

Doc Al
Mentor
F is the horizontal force (same as what the first part of problem asked for)
In the first part, F is the net force on the car, which is just the friction. In the second part, you will use the $\mu_k$ from the first part, but not the F. You are already including friction as well as gravity.
Well, I tried losing the F, which doesn't make much sense to me, and have -ukmgcos(18.9)-mgsin(18.9)=ma --> a=-14.57 m/s^2
What did you use for $\mu_k$?

Doc Al
Mentor
and i found uk with this equation from the first part

5402-ukmg=ma where a=-6.075
uk=1.23
Nope. Your value of force (5402N) is the friction!

ah ok so uk=.619

and equation becomes -(.619)(890)(9.8)cos(18.9) - (890)(9.8)sin(18.9) = ma , solve for a, plug in (27)^2=-2ax

and x=40.89, great thank you!!!!!