A car is moving at 40mph. If the mass of the car is 2,300 kg and the coefficient of kinetic friction is 0.4, what is the stopping distance in ft?
First convert 40mph to m/s
KE = Fn * d
1/2 mv^2 = 0.4mg * d
(v^2)/(2 *0.4*g) = d (in meters, so multiply by 3 to get ft)
is the answer ~120 ft?
Spot on solution, although I came up with a slightly larger final number (in the 140 m +/- range). It comes down to our rounding differences I suspect. In a multiple choice scenario, all three answers so far will likely lead to the best answer of the four choices.