## Stopping distance (work & energy)

1. The problem statement, all variables and given/known data

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 travelling 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

2. Relevant equations

W= F*X
.5mvE2
SumF = ma

3. The attempt at a solution

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

Help anyone!
 Mentor Blog Entries: 1 Show what you tried. Hint: Use the work-energy theorem.
 basically I tried using each of the above and nothing worked I even tried .5mvfE2 -.5mv0E2 - nothing works

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## Stopping distance (work & energy)

Use my hint! (You need another formula that combines your first two.)
 is it potential energy? PE = mgh?
 a bit lost

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 Quote by Chica1975 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?
 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.
 Mentor Blog Entries: 1 What does the work-energy theorem tell you? (See: Work-Energy Principle)