anonymous
  • anonymous
An electric motor is used to lift a mass of 1 kg through a height of 3.2 metres, at a constant speed and in a time of 8 seconds. Calculate the useful power output of the motor.
OCW Scholar - Physics I: Classical Mechanics
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anonymous
  • anonymous
An electric motor is used to lift a mass of 1 kg through a height of 3.2 metres, at a constant speed and in a time of 8 seconds. Calculate the useful power output of the motor.
OCW Scholar - Physics I: Classical Mechanics
chestercat
  • chestercat
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anonymous
  • anonymous
The definition of power is work per unit time... the work is force dot displacement. Calculate the work, divide by the time over which it is accomplished.
anonymous
  • anonymous
yeah did that but I wasnt really sure cos to force =mass * acceleration, but since it is moving at a constant speed, it has no acceleration. I used gravity for the acceleration so wanted to know is thats right.
anonymous
  • anonymous
If you use gravity as the force, then you are calculating the work done by gravity. It will be the negative of the work done by the motor. The cosine term involves the angle between the force vector's direction and the displacement vector, so in this case of gravity it is 180 degrees, and cos(180) is -1, or in the case of the motor, it is 0 degrees and cos(0) is 1.

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