A 6110-kg bus traveling at 20.0 m/s can be stopped in 24.0 seconds by gently applying the brakes.

Question

What average force is exerted on the bus during this stop?

Answer 1

m = 6110 kg, t = 24 sec, v₁ = 20 m/sec a = average acceleration = (v_f - v₁)/(t) = (0-20)/24 = -5/6 m/sec2

so, F = ma = (6110kg)(-5/6) = 5,091.7 kg-m/sec² = 5,091.7 N

Another method: Work = F* Distance = Change in Kinetic Energy

the distance x = x₁ + at²/2 = 0 + (.5)(-5/6)(24²) = 240 m

Change in KE = mv₁²/2 = (6110)(20)² = 1,222,000 = F*D = 240*F

F = 1,222,000/240 = 5,091.7 N