Can You Solve This Physics Problem on Kinetic and Potential Energy?

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Homework Help Overview

The discussion revolves around a physics problem involving kinetic and potential energy, specifically focusing on the dynamics of two masses connected by a rope on an inclined plane. Participants are attempting to solve parts (c), (d), and (e) of the problem, which require calculations related to energy changes as one mass falls and the other rises.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are sharing their calculations for potential energy changes and questioning the correctness of their results. There is an exploration of the relationship between kinetic and potential energy, particularly in a frictionless system. Some participants express confusion over previous answers and seek clarification on how to relate different parts of the problem.

Discussion Status

The discussion is active, with participants revisiting their calculations and seeking guidance on how to approach the remaining parts of the problem. There is a recognition of previous errors, and some participants are beginning to connect the concepts of energy conservation and the relationships between kinetic and potential energy.

Contextual Notes

Participants are working under the assumption that the inclined plane is frictionless, which influences their calculations and reasoning. There is also a sense of urgency expressed by some participants regarding the need for clarification and assistance in solving the problem.

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c, d, and e I don't know how to solve

For d I tried doing

vf2 = vi2 +2ad
Vf = √(2*9.8*2.3) = 6.71 m/s

Assuming the inclined plane is frictionless
Since m2 is roped to m1, it's velocity up the ramp will be 6.71m/s too

The total increase in KE is then 0.5[8+6]*6.71^2 = 315.16 J

^

"INCORRECT"

Right now I need a genius to tell me how to get the answer ASAP -.-

Since (d) is wrong, so is e. -.-
Story of my life.
 
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riseofphoenix said:
Help.png


c, d, and e I don't know how to solve

For d I tried doing

vf2 = vi2 +2ad
Vf = √(2*9.8*2.3) = 6.71 m/s

Assuming the inclined plane is frictionless
Since m2 is roped to m1, it's velocity up the ramp will be 6.71m/s too

The total increase in KE is then 0.5[8+6]*6.71^2 = 315.16 J

^

"INCORRECT"

Right now I need a genius to tell me how to get the answer ASAP -.-

Since (d) is wrong, so is e. -.-
Story of my life.

For (c), how did you get that number? Isn't it just the sum of the two ΔU values?
 


berkeman said:
For (c), how did you get that number? Isn't it just the sum of the two ΔU values?

m1 falls 2.3m and therefore has -6*9.8*2.3 = -135.24 J difference.

m2's PE increases 8*9.8*sin 30° = +39.2 J difference.

The total change then is 39.2 - 135.24 = -96.04 J

^ That was my previous answers, but I got that wrong too...
So I'm pretty much screwed.
 


riseofphoenix said:
m1 falls 2.3m and therefore has -6*9.8*2.3 = -135.24 J difference.

m2's PE increases 8*9.8*sin 30° = +39.2 J difference.

The total change then is 39.2 - 135.24 = -96.04 J

^ That was my previous answers, but I got that wrong too...
So I'm pretty much screwed.

You left off the 2.3 for m2...
 


berkeman said:
You left off the 2.3 for m2...

(c)

m1 falls 2.3m and therefore has -6*9.8*2.3 = -135.24 J difference.

m2's PE increases 8*9.8*sin 30°*2.3 = +90.16 J difference.

The total change then is 90.16 - 135.24 = -45.08 J

Ok so how does this relate to part d and e though? How would I solve for those two last ones?

(d)

(e)
 


riseofphoenix said:
(c)

m1 falls 2.3m and therefore has -6*9.8*2.3 = -135.24 J difference.

m2's PE increases 8*9.8*sin 30°*2.3 = +90.16 J difference.

The total change then is 90.16 - 135.24 = -45.08 J

Ok so how does this relate to part d and e though? How would I solve for those two last ones?

(d)

(e)

(d) asks for the change in KE. Since the system is frictionless...
 


berkeman said:
(d) asks for the change in KE. Since the system is frictionless...

But I already tried solving for it and i got it wrong... (??)
 


riseofphoenix said:
But I already tried solving for it and i got it wrong... (??)

But you had the wrong answer for (c) before. Now you have that right.

What is the total energy of a system in terms of the potential and kinetic energies? If the system is lossless (no friction), what can you say about the total energy over time...?
 


berkeman said:
But you had the wrong answer for (c) before. Now you have that right.

What is the total energy of a system in terms of the potential and kinetic energies? If the system is lossless (no friction), what can you say about the total energy over time...?

KEinitial + PEinitial = KEfinal + PEfinal
KEinitial + PEinitial = 0 + 0 (No friction)
KEinitial + PEinitial = 0?
 
  • #10


riseofphoenix said:
KEinitial + PEinitial = KEfinal + PEfinal
KEinitial + PEinitial = 0 + 0 (No friction)
KEinitial + PEinitial = 0?

Rearrange so you show the relationship between ΔKE and ΔPE...
 
  • #11


KE = -PE
KE = -mv?
 
  • #12


riseofphoenix said:
KE = -PE
KE = -mv?

If you put Δs in front of your first equation, you would be on the right track...
 
  • #13


Consider the total energy over time which is a constant. Hence change in PE must balance out change in KE. The rest is obvious.
 
  • #14


As for (e) consider that you have a final change in KE, as well as the masses. You can now plug that into solve for velocity.
 
  • #15


Oops, Looks like riseofphoenix is on a temporary vacation from the PF for a different thread.
 

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