Wc = m*g = 1840kg * 9.8N/kg = 18032 N. =
Wt. of car.
Fc = 18,032N[12.5o] = Force of the car.
Fp = 18,032*sin12.5 = 3903 N. = Force
parallel to incline.
F-Fp = m*a = m*0 = 0
F-3903 = 0
F = 3903 N. Parallel to the incline
Work = F*d = 3903 * 14 = 54,642 J.
What is the minimum work in joules needed to push a 1840 kg car 14.0 m up a 12.5 degree incline? (Ignore friction.)
1 answer