- #1
darkwolfe5
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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 [tex]\theta[/tex] * v
970*(9.8 sin 30) * 2.2 = 1.05*10^4
b) P[tex]_{}max[/tex] = m * (v/t + g sin [tex]\theta[/tex]) * 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 [tex]\theta[/tex])*d
970(2.2/10.5 - 9.8 sin 30) * 1400 = 6.8*10^6
or: W = m*g*d*sin [tex]\theta[/tex])
970(9.8)(1400)(sin 30) = 6.65*10^6
or: W = m*g*sin [tex]\theta[/tex])*(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)