Power and maximum power problem

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Homework Help Overview

The problem involves a loaded ore car with a mass of 950 kg being pulled up an inclined mine shaft at 30 degrees. The car accelerates to a speed of 2.2 m/s over 12 seconds and then moves at a constant speed. The questions focus on calculating the power required by the winch motor during constant speed and the maximum power needed.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between force, power, and velocity, with some attempting to apply the formula P = F * V. There is confusion regarding the difference between the two parts of the problem, particularly what changes when calculating maximum power. Questions arise about the forces acting on the car during acceleration versus constant speed.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the importance of free body diagrams and the correct application of force equations. However, there is no explicit consensus on the approach to part (b) of the problem.

Contextual Notes

Participants note the need to consider both gravitational forces and the forces due to acceleration when calculating the required power. There is also mention of potential errors in calculations and the importance of correctly summing 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
 
Last edited:
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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