Power and maximum power problem

AI Thread Summary
The discussion centers on calculating the power requirements for a winch motor pulling a loaded ore car up an inclined mine shaft. For part (a), the power needed at constant speed is calculated using the formula P = F * V, resulting in approximately 10.2 kW. Part (b) requires understanding that maximum power occurs during acceleration, necessitating the use of net force calculations that account for both gravitational force and acceleration. The correct approach involves using the equation T - mgsin(30) = ma to find the tension in the cable. Emphasis is placed on the importance of free body diagrams for accurately solving physics problems related to forces.
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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 don't know what to do
 
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I messed up on the last line... i meant i don't know how to do (b) but i think its like (a)
can anyone 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 I am not sure about but what would it change to?? the sin I am certain would stay the same aswell. I don't really understand the difference between part a and b
 
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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
 
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 :)
 
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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