Power involved in running upstairs

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A 65 kg student jogs up a vertical distance of 17.6 m in 65 seconds, requiring the calculation of power expended against gravity. The key equations involve force (F = m*A), work (W = F*d), and power (P = W/t). The student gains potential energy as they ascend, which is crucial for understanding the work done against gravity. The discussion emphasizes the importance of incorporating gravitational force into the calculations to find the correct power output. Ultimately, the solution was found to be simpler than initially expected.
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Homework Statement



A 65 kg student jogs upstairs from the first floor to the sixth, a vertical distance of 17.6 m, in 65 s. Find the power the student expends working against gravity (In kW.)

Homework Equations



F=m*A
W=F*d
P=W/t

The Attempt at a Solution



Tried solving here but didn't know how to incorporate gravity into this, got a crazy answer.
 
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So power is in Watts, which is Joules per Second. What does the student gain when (s)he runs upstairs?
 
The student would gain height as they climbed.
 
But why is it hard work going upstairs, but not down them?
 
Because gravity is weighing down on you as you ascend. I understand it physically, but I don't know where to incorporate g mathematically.
 
If you lift a weight up into the air, what (other than height) are you giving it? If you drop it, it gains kinetic energy, but where does that energy come from?

Sorry that it appears I'm taking a really roundabout route with this; I'm trying to get you to come up with the answer!
 
It gains potential energy as it gains height, so can I use the law of conserved forces?
 
Excellent, yes it does. So you should be able to calculate how much energy the student has gained, and you know how long it took...
 
Got the right answer, thank you. That was a lot simpler than I thought it was... lol
 
  • #10
These things often are!
 
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