Question

A 1000 kg car traveling at a speed of 40 m/s skids to a halt on wet concrete where the coefficient of friction=0.60. How long are the skid marks?

Answer 1

F = ma
a = F(friction) / m

F(friction) = mu F(normal)
F(friction) = mu g m
a = mu g m / m
a = mu g
a = 0.6 * 9.81m/s^2
a = 5.886 m/s^2

v^2 = 2 a x
x = v^2 / 2 a
x = (40m/s)^2 / 2 * 5.886m/s^2
x = 136m

Answer 2

Force

F = mg = (1000 kg )( 9.8 m/s^2) = 9800 N.
braking force.

   F_b = mu mg = ( 0.6 ) ( 9800 ) = 5880 N

Acceleration = (f/m) = 5880 / 1000 = 5.88m/sec^2.
Distance = (v^2/2a), = ( 40 m/s )^2 / 2 (5.88 ) =136 m.

Answer 3