A 1100 kg car traveling at a speed of 39.0 m/s skids to a halt on wet concrete where mu_k = 0.680?

Question

A 1100 kg car traveling at a speed of 39.0 m/s skids to a halt on wet concrete where mu_k = 0.680 .How long are the skid marks?

Answer 1

f=uN
f=(.68)(1100)(-9.8)
f=7330.4
F=ma
-7330.4=(1100)a
a=-6.664
vf^2=vi^2+2as
0^2=39^2+2(-6.664)s
s=114.12m

Answer 2

Work done by forcefriction = change in kinetic energy of car

So mu_k*m*g*distance = 1/2 m*v^2

Cancel m and solve for distance

Answer 3

known variables:

v(initial) = 39 m/s
v(final) = 0 m/s
a = 0.680g; because Fnet = ma = F(friction) = (mu_k)(m)(g)

using the kinematic equation v^2 = vinitial^2 +2ad;

d = 39^2 / (2 x 0.680g) = 1521 / 13.32 = 114 m

Answer 4

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Answer 5

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