Question
A truck with a mass of 1.5 x 103 kg accelerates to a speed of 18.0 m/s in 12.0 s from a dead stop. Assume that the force of resistance is a constant 400.0 N during the acceleration. What is the average power developed by the truck’s engine?
Answers
a = (V-Vo)/t = (18-0)/12 = 1.5 m/s^2.
Wt = m*g = 1500kg * 9.8N/kg = 14,700 N.
d = 0.5a*t^2 = 0.75*12^2 = 108 m.
Fe-400 = m*a
Fe-400 = 1500*1.5 = 2250
Fe = 2250 + 400 = 2650 N. = Force of
engine.
Power=Fe * d/t=2650 * 108/12=23,850 J/s
= 23.850 Watts.
Wt = m*g = 1500kg * 9.8N/kg = 14,700 N.
d = 0.5a*t^2 = 0.75*12^2 = 108 m.
Fe-400 = m*a
Fe-400 = 1500*1.5 = 2250
Fe = 2250 + 400 = 2650 N. = Force of
engine.
Power=Fe * d/t=2650 * 108/12=23,850 J/s
= 23.850 Watts.