Asked by falcon306

A 2000 {\rm kg} car traveling at a speed of 13 {\rm m/s} skids to a halt on wet concrete where mu_k = 0.50. How long are the skid marks?

I tried doing this many times and each answer is wrong. Can someone help?
Answered by Henry
Fc = mg = 2kg * 9.8N/kg = 19.6N = Force
of car.

Fc = 19.8N @ 0 deg.(Hor. plane).

Fp = mg*sin(0) = 19.8sin(0) = 0 = Force
parallel with plane.

Fv = 19.6cos(0) = 19.6N = Force perpendicular to plane.

Ff = u*Fv = 0.5 * 19.6 = 9.8N.

Fn = Fp - Ff = 0 - 9.8 = -9.8N = Net
force.

Fn = ma,
a = Fn/m = -9.8/2 = -4.9m/s^2.

Vf^2 = Vo^2 + 2ad,
d = (Vf^2-Vo^2) / 2a,
d = (0-(13)^2 / -9.8 = 17.2m. = Length
of skid marks.
Answered by Henry
Correction:
mass of the car = 2000kg: NOT 2kg.

Fc = 2000kg * 9.8N/kg = 19,600N.

Fp = 19,600sin(0) = 0.

Fv = 19,600cos(0) = 19,600N.

Ff = u*Fv = 0.5 * 19,600 = 9800N.

Fn = Fp - Ff = o - 9800 = -9800N.

a = Fn/m = -9800 / 2000 = -4.9m/s^2.

The remaining calculations are correct
as is.

Answered by falcon306
Thank you very much!
Answered by Henry
Glad i could help!
Answered by CANTIUS
work = change in kinetic energy
friction force X distance = change in KE
-mu_k n x = 0 - 1/2 m v^2
[Remember, friction does negative work)
mu_k m g x = 1/2 m v^2
Simplifying:
x = v^2/(2 mu_k g) = 17.24 m

Notice that in this solution, the mass of the car is not needed.
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