Gravitational potential energy increase

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Homework Help Overview

The discussion revolves around a problem related to gravitational potential energy, specifically calculating the increase in potential energy for a person riding an elevator. The context includes the weight of the person, their speed, and the duration of the ride.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the potential energy formula (PE = mgh) and question the relationship between potential energy and kinetic energy. There are attempts to clarify the meaning of variables and the correct application of equations.

Discussion Status

Some participants have offered guidance on focusing on the potential energy equation and solving for displacement. There is acknowledgment of confusion regarding the application of different energy equations, but no consensus has been reached on the correct approach yet.

Contextual Notes

Participants are grappling with the definitions of variables and the appropriate equations to use. There is a noted misunderstanding in equating potential energy with kinetic energy in this context.

majormuss
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Homework Statement



A person weighing 6.0 × 102 Newtons rides an
elevator upward at an average speed of 3.0 meters
per second for 5.0 seconds. How much does this
person’s gravitational potential energy increase
as a result of this ride?
(1) 3.6 × 102 J (3) 3.0 × 103 J
(2) 1.8 × 103 J (4) 9.0 × 103 J

Homework Equations




PE=mgh, KE=1/2mx^2

The Attempt at a Solution


I tried solving the question this way but it seems am wrong... mgh=1/2mx^2 and I arrived at number '3'. Need help!
 
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U = mgh ... that is about all you'll need (you know m and g, you just need to solve for the displacement).
 
Gear300 said:
U = mgh ... that is about all you'll need (you know m and g, you just need to solve for the displacement).

wat is U?
 
wat is U?
 
majormuss said:
wat is U?
Potential Energy=mass*g*height

majormuss said:
I tried solving the question this way but it seems am wrong... mgh=1/2mx^2 and I arrived at number '3'. Need help!

Your approach can't be used in this type of problem. If you wanted to find the speed of a car at the bottom of a hill, you can set the two equations equal to each other (PE for the top, KE, for the bottom).

A person weighing 6.0 × 102 Newtons rides an
elevator upward at an average speed of 3.0 meters
per second for 5.0 seconds.

Try finding the three parts of the equation Gear300 posted...you just need to solve for height (or displacement).
 
thanx a bunch, just realized my blunder...
 

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