A 700 kg Toyota collides into the rear end of a 2400 kg Cadillac stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 5.0 m before stopping. The police officer, knowing that the coefficient of kinetic friction between tires and road is 0.40, calculates the speed of the Toyota at impact. What was that speed?
2007-02-21 10:44:11 · 1 answers · asked by Eddie G 1 in Science & Mathematics ➔ Physics
F = kN = force of friction on the two locked cars = k(m + M)g; where k = .4, m = 700 kg, M = 2400 kg, and g = 9.81 m/sec^2.
W = KE = Fd = kNd = 1/2 mv^2; where the kinetic energy of the Toy is expended by the work (W) when the locked cars are moved d = 5 m. Thus, v^2 = 2kNd/m = 2k(M + m)gd/m; so that, v = sqrt(2kdg(M + m)/m) = sqrt(2*.4*5*9.81*(3100/700)) and you can do the math.
Lesson learned: The kinetic energy provided by the moving Toy is expended by working to move the combined masses of the two cars a distance d against the force of friction. QUES: where do you suppose that kinetic energy went to...remember energy can be neither created nor destroyed...so where did KE go?
2007-02-21 11:34:45 · answer #1 · answered by oldprof 7 · 0⤊ 0⤋