What is the minimum work needed to push a 950 kg car 890 m up along a 8.0 degree incline? Assume the effective coefficient of friction retarding the car is 0.22.

Answer 1

Wc = mg = 950kg * 9.8N/kg 9310N. = Weight of car.

Fc = (9310N,8deg.) = Force of car.

Fp = 9310sin8 = 130N. = Force parallel to ramp.

Fv = 9310cos8 = 9219N = Force perpendicular to ramp = The normaql.

Ff = u*Fv = 0.22 * 9219 = 2028N. = Force of friction.

Fn = Fap - Fp - Ff = ma = 0. a = 0.
Fap - 130 - 2028 = 0,
Fap = 130 + 2028 = 2158N = Force applied.

W = F*d = 2158 * 890 = 1,920,620J.