Why didn't this work? (Pun intended)

[Kyle.Nemeth]

Homework Statement

A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mineshaft by a cable connected to a winch. The shaft is inclined at 30^o above the horizontal. The car accelerates uniformly to a speed of 2.20 m/s in 12.0 s and then continues at constant speed. (a) What power must the winch motor provide when the car is moving at constant speed? (b) What maximum power must the winch motor provide? (c) What total energy has transferred out of the winch motor by work by the time the car moves off the end of the track, which is of length 1250 m?

Homework Equations

P=W/t
W=ΔK+ΔU
v=v_o+at
x=x_o+v_ot+½at²
Sinθ=O/H

The Attempt at a Solution

I'm okay with (a), but I'm working on part (b). I assume that the max power the winch must provide can be found by considering the total change in mechanical energy of the system over a time t.

My result is,

P_max=½mv[(v/t)+gsinθ]

This gives me half of the max power. Does the total change in mechanical energy of the system not govern the max power output of the winch?

Does the net force on the cart govern the max power the winch must provide?

 

[TSny]

Question (b) is asking for the maximum instantaneous power. Do you know how to express the instantaneous power in terms of force F and velocity v?

 

[Kyle.Nemeth]

Yeah,

P=F⋅v

right?..

I used P=Fv, then I used Newtons second law to show that,

P_max=mv[(v/t)+gsinθ]

Which is the correct answer.. I'm wondering why this approach has given me the right answer but my approach gave me half of the correct answer.

 

[TSny]

Total change in energy over time gives average power for the interval of time rather than an instantaneous power.

At the start, t = 0, the power is zero since P = Fv and v = 0 at t = 0.
At time t_f = 12.0s just before the acceleration goes to zero the instantaneous power is P = Fv_f where v_f is the velocity at time t_f. P = Fv_f is the maximum instantaneous power.

Due to the fact that the acceleration is constant between t = 0 and t = t_f, the average power is just the average of the initial and final power:

P_avg = (1/2)(0 + Fv_f) = (1/2)Fv_f = (1/2)P_max.

 

[Kyle.Nemeth]

Ahhh, okay. I understand.

Thank you for your help. Much appreciated