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Stopping distance (work & energy)

  • Thread starter Chica1975
  • Start date
  • #1
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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 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

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!
 

Answers and Replies

  • #2
Doc Al
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44,910
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Show what you tried. Hint: Use the work-energy theorem.
 
  • #3
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basically I tried using each of the above and nothing worked I even tried .5mvfE2 -.5mv0E2 - nothing works
 
  • #4
Doc Al
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44,910
1,169
Use my hint! (You need another formula that combines your first two.)
 
  • #5
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is it potential energy? PE = mgh?
 
  • #6
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a bit lost
 
  • #7
Doc Al
Mentor
44,910
1,169
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?
 
  • #8
63
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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.
 

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