A 1850-kg car experiences a combined force?
A 1850-kg car experiences a combined force of air resistance and friction that has the same magnitude whether the car goes up or down a hill at 26 m/s. Going up a hill, the car's engine produces 43 hp more power to sustain the constant velocity than it does going down the same hill. At what angle is the hill inclined above the horizontal?
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Power developed(J/s) = net force (Fn) x velocity(v)
Going uphill the net opposing force, Fn = weight component (mg sinθ) + friction (f)
Fn = mg sinθ + f
Power up .. Pu = Fn x v .. Pu = [mgv sinθ + fv]
Going down hill the net opposing force, Fn' = f - mg sinθ .. (assuming f > mg sinθ)
Fn' = f - mg sinθ
Power down .. Pd = Fn' x v .. [Pd = fv - mgv sinθ]
Pu = Pd + 3.20^4 J/s(43 hp)
mgv sinθ + fv = fv - mgv sinθ + 3.20^4
2mgv sinθ = 3.20^4 ..
sinθ = 3.20^4 / 2mgv = 3.20^4 / (2 x 1850kg x 9.80 x 26m/s) .. sinθ= 0.0339
.. ►θ = 1.95º
that's common. allow's evaluate the flow up a hill. Engine produces the stress F1 = Ffriction + Fresistance + mg sin a. the flexibility of the engine is P1 = F1 * V at the same time as the motor vehicle strikes down the stress is F2 = Ffriction + Fresistance - mg sin a. P2 = F2* V P1-P2 = F1*V - F2*V = (F1-F2)V = 2 mg * sin a * V and ultimately sin a = (P1-P2)/(2 mg * V) i'm valuable you may calculate that your self. sturdy luck!