Help Cure High Schooler's Anxiety: Solve 12 Energy Questions!

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Homework Help Overview

The discussion revolves around a set of twelve energy-related questions, primarily focusing on concepts of kinetic energy, potential energy, and work in various physical scenarios. The original poster expresses difficulty in understanding energy transformations and seeks assistance in solving the problems presented.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants question the correctness of specific problem statements and the assumptions underlying the energy calculations. There are discussions about the application of energy conservation principles, the relationship between mass and kinetic energy, and the interpretation of potential energy in various contexts.

Discussion Status

The conversation includes attempts to clarify misunderstandings and provide hints without directly solving the problems. Some participants emphasize the need for the original poster to demonstrate effort and understanding of the concepts involved. There is an ongoing exploration of different interpretations of the questions, particularly regarding energy transformations and the relationships between kinetic and potential energy.

Contextual Notes

Participants note that the original poster may need to adhere to specific homework guidelines, which could limit the amount of assistance provided. There is also mention of varying assumptions, such as the value of gravitational acceleration used in calculations.

  • #31


Oh god. I got a 95 on the kinematics test, studied forever hoping that this stuff wouldn't come back to haunt me... and now it has. We just started the work/energy/optics unit. I want to understand this.

So from deriving several equations I got v = deltax/sqrt(deltay/.5g)

Assuming that the initial horiz. speed was 5.9 m/s, I got deltay = .45m
 
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  • #32


My answer is closest to .4 m. Do you think that's right?
 
  • #33


Huh, interesting.

Try this

x = vt (x is horizontal distance)

y = -gt^2/2 + 2 (vertical distance, check that t = 0, y = 2)

There is no velocity before because it is freefalling.

Your answer is wrong, because the base of the ramp is 2m high, why would it be 0.4m...? But I am not really reading your posts so I am not much of help, but I have set up some eqtns for you
 
  • #34


I mean 2.4 m because you must add 2 to the .4m.
 
  • #35


See, my derived equation allows me to neglect time! :D
I found it by myself... so proud.

However, I think I neglected to add 2 m to .4 m because it asks for with respect to the floor. Look:

8. A marble rolls down a ramp that sits on table. The marble leaves the base of the ramp with a horizontal velocity. The base of the ramp is 2 meters above the floor. The marble strikes the floor at a distance of 1.8 meters from where it left the ramp. Give the height of the ramp with respect to the floor.
Choose one answer.
2.4 m
0.4 m
4.4 m
1.4 m
2.2 m
 
  • #36


Yup, so remember they want the height from the floor so the answer should be 2.4m
 

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