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abot
Nov19-04, 04:54 PM
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

abot
Nov19-04, 05:10 PM
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

abot
Nov19-04, 06:49 PM
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.

abot
Nov19-04, 08:01 PM
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