Power problem - finding altitude change

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Homework Help Overview

The problem involves calculating the altitude change achievable by an electric scooter given its energy capacity and the weight of the rider and scooter. The context includes energy losses due to friction and other factors.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate power and energy to altitude change using the equation Ug = mgh but arrives at an incorrect value. Some participants question the definition of watt-hours and seek clarification on its implications for the problem. Others suggest converting watt-hours to joules and exploring the relationship between power and energy.

Discussion Status

Participants are actively engaging with the problem, seeking clarification on units and concepts. Some have shared their attempts at solving related aspects, while others are encouraged to elaborate on their reasoning to facilitate further assistance.

Contextual Notes

There is a focus on understanding the implications of energy losses and the definitions of units involved, which may affect the approach to the problem. Participants are encouraged to share their thought processes and attempts to foster collaborative problem-solving.

apang42
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Homework Statement



An electric scooter has a battery capable of supplying 140 Wh of energy. If friction forces and other losses account for 60.0% of the energy usage, what altitude change can a rider achieve when driving in hilly terrain, if the rider and scooter have a combined weight of 860 N?

Homework Equations



Ubattery= 140(.6)/3600s = .015556W
Ug= mgh

The Attempt at a Solution



.015556 = 860h
h = 1.8E-5
this is incorrect.

can anyone point me in the right direction? thanks for your help!
 
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Units: What is the definition of a Wh?
 
Wh = watt hours
 
But what does "Watt hours" mean? What does "140 Watt hours" mean? Can you explain that in words? It will help you with the problem. (If you're not sure, search for the definition of a kWh, a kilowatt hour.)
 
you can get J out of it right? Then since you have J there is some equation to get m. maybe mgh? I don't know I need help on this problem too.

Thanks,
C
 
csg39 - in reply to your private message, please post the problem in the forum, including your attempts at a solution so far, where everyone can see and help.
 
its the same problem as above.
 
so i solved for velocity by using P=Fv=(mg)v
 
csg39 said:
so i solved for velocity by using P=Fv=(mg)v
You'd need to explain how you did that - what you wrote down and what your thought process was - for people to be able to help you on here.

My other tip, if you're still stuck, would be to have a go at answering the question in post 4 of this thread.
 
  • #10
Hw can U equate power with mgh i.e energy?
 

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