What Are the Equations for Power and Tractive Effort in a Car Ascending a Slope?

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To calculate the power developed by a car ascending a slope with friction, one must consider both the work done against gravitational forces and rolling resistance. The power at 30 km/h is determined to be 11.4 kW, based on the car's mass, acceleration, and the incline's angle. The tractive effort involves calculating the total force required to overcome both gravitational pull and friction, which remains unresolved in the discussion. Understanding D'Alembert's principle is crucial for deriving these equations accurately. The conversation emphasizes the importance of breaking down the forces involved to find the correct equations for power and tractive effort.
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Hi
I need a bit of help with D'Alembert's principle, I'm after the equations to find the power in a car ascending a slope with fritional resistance. The example I will give is not my assignment question as I want to work that out for myself, but an example from a textbook that just gives me the answers and not the equations, but I cannot find a way using KE and PE to get it right.

A car of mass 800kg accelerates from rest to a speed of 50km/h in 10 seconds, while ascending an incline of 1 in 70. The rolling resistance is 150N. Assuming uniform acceleration, what is

1 the power developed at 30 km/h and
2 the tractive effort

Answer to number 1 is 11.4KW, number 2 is another baffling question I’m afraid.

I have worked out that the acceleration is 1.389m/s and the distance traveled is 13.89

Any help gratefully received
 
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Welcome to PF mitch,

Power is simply the rate at which work is done. Therefore, if one could find the work done then calculation of the power would be trivial.
 
Last edited:
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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