Potential Energy Change

  • Thread starter winterma
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A .4 kg mass moves up 40 cm along an incline. The vertical height of the incline is 7.0 cm. The potential energy change is??

I used the formula for PEC. 1/2k(deltaX)^2 which got me to 320 joules. That is wrong though. What am i doing wrong???

If a force of .85N pulled parallel up along the surface of the incline is required to raise the mass back to the top of the incline, how much work is done?

i used W=fd
w=.85(40)
w=34

but that answer is wrong. Im doing something wrong.
 

Answers and Replies

  • #2
AKG
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winterma said:
A .4 kg mass moves up 40 cm along an incline. The vertical height of the incline is 7.0 cm. The potential energy change is??

I used the formula for PEC. 1/2k(deltaX)^2 which got me to 320 joules. That is wrong though. What am i doing wrong???
What's "PEC"? And if I'm not mistaken, the expression (not formula) you've given is that for the potential energy due to the extension of a spring -- this problem has nothing to do with springs. Gravitational potential energy, in this case, is given by:

[tex]\Delta E_{grav} = mg\Delta h[/tex]

Where m is the mass of the object, g is the gravitational constant, and [itex]\Delta h[/itex] is the change in height. A little trigonometry will allow you to determine the total change in height of the object.

If a force of .85N pulled parallel up along the surface of the incline is required to raise the mass back to the top of the incline, how much work is done?

i used W=fd
w=.85(40)
w=34

but that answer is wrong. Im doing something wrong.
At this point, I've given you a number of formulas, but the questions you ask suggest that you aren't understanding some of the basic concepts of these physics questions. You can find the formulas yourself, and you should, yourself, be able to figure out which formulas to use, and when, but you don't seem to get the concepts. Make sure you take time to understand these problems and whatever your textbook says, even though you can quickly find the right formula and plug in the right numbers. Now, tell us what your books says the answer is, and maybe we'll try and work backwards.
 
  • #3
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There is something goofy about the question. In the first part, does the mass go up or down the incline?

It makes no difference, because you have a mass of 0.4 kg that changes elevation by 7 cm. Therefore the mass will lose or gain (depending on the direction of travel) PE = mgh = 0.4(9.8)(0.07) = ... joules of potential energy during the process.
 
  • #4
so rounded off, the G.P.E is 0.28 joules. i dont see why you require harder formulas than just gpe = mgh.
 

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