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A loaded ore car has a mass of 950kg 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 degrees above the horizontal. the car accelerates uniformly to a speed of 2.2m/s in 12 sec and continues at constant speed. (a) what power must the winch motor proide when the car is moving at constant speed ? (b)what maximum power must the winch motor provide?
I know (A) is....
P= F*V
P=mgsin30 * V
P=10.2Kw
Know how do you do (b) i think its similar to (a) i dont know what to do
I messed up on the last line... i meant i dont know how to do (b) but i think its like (a)
can any one shed some light
ohhhh man! does anyone have any idea???? haha
if its asking for the maximum power what would change. the mass is the same so is g the velocity im not sure about but what would it change to?? the sin im certain would stay the same aswell. I dont really understand the differene between part a and b
thermodynamicaldude
Nov19-04, 07:26 PM
F = ma
Thus, when the car is accelerating, the force acting on the car not only must overcome the force of gravity, but it also must accelerate the car.
what??? well if you find the F that way then....
F=ma (1st i need acceleration.....(2.20m/s / 12sec)= .183m/s^2
F=950kg*.183m/s^2 F=174N
So then...
P=FV
P=174sin30*2.2 P=191W
but the answer is P=10600W
Spectre5
Nov19-04, 08:19 PM
Watch what you do....
it is not F=ma
it is Sum(F) = ma
YOu have the cable pulling in one direction and the weight coming down in another...
thuse your eq is:
T - mgsin(30) = ma
Where T is the tension and thus the force in the cable
Make sure you look at the free body diagram :)
thermodynamicaldude
Nov19-04, 10:14 PM
yeah...sorry 'bout that..I left out the sumation part... but basically follow Spectre5's advice...free body diagrams are essential to solving most force related Physics problems
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