Stopping distance (work & energy)

[Chica1975]

Homework Statement

in an emergency stop on a level dry concrete road the magnitude of the friction force when sliding is approx 80% of the weight of the vehicle. What stopping distance is required for a vehicle traveling at 88km/h (24.444 m.s)?

Assume that all the wheels lock when the brakes are applied

Use 9.8 m/s2 for gravitational acceleration

Homework Equations

W= F*X
.5mvE2
SumF = ma

The Attempt at a Solution

I have tried a number of combinations using the above equations - I can't figure this out.

Help anyone!

 

[Doc Al]

Show what you tried. Hint: Use the work-energy theorem.

 

[Chica1975]

basically I tried using each of the above and nothing worked I even tried .5mvfE2 -.5mv0E2 - nothing works

 

[Doc Al]

Use my hint! (You need another formula that combines your first two.)

 

[Chica1975]

is it potential energy? PE = mgh?

 

[Chica1975]

a bit lost

 

[Doc Al]

is it potential energy? PE = mgh?

No. Gravitational PE is not relevant here since the height doesn't change.

How does the initial KE of the car relate to the work done by friction in stopping the car?

 

[Chica1975]

to be honest I don't know - it must reduce kinetic energy or change it in some way becoz friction is going in the opposite direction?

I am completely lost.

 

[Doc Al]

What does the work-energy theorem tell you? (See: http://hyperphysics.phy-astr.gsu.edu/hbase/work.html#wepr")