What total energy transfers from this motor during this work?

[darkwolfe5]

Homework Statement

A loaded ore car has a mass of 970 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 30.0° above the horizontal. The car accelerates uniformly to a speed of 2.20 m/s in 10.5 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 transfers out of the motor by work by the time the car moves off the end of the track, which is of length 1400 m?

Homework Equations

W=F*d
P = F*v = W/t

The Attempt at a Solution

I've gotten the answers to a & b
a) P = m * g * sin $$\theta$$ * v
970*(9.8 sin 30) * 2.2 = 1.05*10^4
b) P$$_{}max$$ = m * (v/t + g sin $$\theta$$) * v
970 * (2.2/10.5 + 9.8) * 2.2 = 1.09*10^4

Now comes the part I don't understand

c) total energy should be the (energy of the acceleration phase) + (energy of constant velocity phase) so right now I have:
W = m*(a - g*sin $$\theta$$)*d
970(2.2/10.5 - 9.8 sin 30) * 1400 = 6.8*10^6

or: W = m*g*d*sin $$\theta$$)
970(9.8)(1400)(sin 30) = 6.65*10^6

or: W = m*g*sin $$\theta$$)*(1400-11.6) + 5428*11.6) = 6.66*10^6
11.6 being the distance traveled during acceleration

None of these is accepted as the correct answer I keep getting the response "your answer differs by order of magnitude" (REALLY wish I knew what that meant)

 

[Nate810]

and the correct answer is 5.6*10^6I'm guessing I'm missing something but I can't figure out what it is. Any help would be appreciated!

 

[kerimek]

Can someone please help me with this?

I would first like to clarify a few things. It seems like you have already solved parts a and b correctly, so I will focus on part c. It is important to note that the total energy transferred from the motor during this work is equal to the total work done on the car, which is equal to the change in the car's kinetic energy.

In this problem, the car starts from rest and accelerates to a constant speed, so we can break down the total energy transfer into two parts: the energy transferred during the acceleration phase and the energy transferred during the constant velocity phase.

During the acceleration phase, the car is gaining kinetic energy, which is equal to 1/2*m*v^2. So the work done on the car during this phase is equal to the change in kinetic energy, which is:

W_1 = 1/2*m*(v_final^2 - v_initial^2) = 1/2*970*(2.2^2 - 0^2) = 1.1*10^3 J

During the constant velocity phase, the car maintains its speed, so there is no change in kinetic energy. Therefore, the work done on the car during this phase is equal to zero.

The total energy transferred from the motor is then:

W_total = W_1 + W_2 = 1.1*10^3 J + 0 J = 1.1*10^3 J

This is the total energy transferred from the motor during this work. I hope this helps clarify any confusion you may have had.