Work and Energy Theorem and Kinetic Energy

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Homework Help Overview

The discussion revolves around the application of the Work and Energy Theorem in the context of a space probe's kinetic energy and the forces acting on it. The original poster presents a problem involving the calculation of the final speed of a space probe after applying a force over a distance.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the Work and Energy Theorem by calculating work done and relating it to the change in kinetic energy. Participants question the manipulation of the equation and suggest reviewing the steps taken in the calculations.

Discussion Status

Participants are actively engaging with the original poster's calculations, with one suggesting a review of the steps to identify potential errors. There is an acknowledgment of a possible typo in the energy value, which may affect the calculations.

Contextual Notes

The original poster mentions that their answer selections seem odd, indicating potential confusion or misalignment with expected results. There is a focus on ensuring unit consistency and addressing any assumptions made during the problem-solving process.

rmarkatos
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A 5.0x10^4kg space probe is traveling at a speed of 11,000m/s through deep space. Retrorockets are fired along the line of motion to reduce the probe's speed. The retrorockets generate a force of 4.0x10^5N over a distance of 2500km. What is the final speed of the probe.

For this problem i know exactly what to do but i know i am getting the wrong answer because I have odd answer selections.

I used W=Fd=1/2m(vf^2)-1/2m(vi^2). I know this is correct because when i plug in the correct answer for vf=9,000 m/s it equals the other side of the equation so i know i am right but for some reason in the process of solving i am not sure what i am doing wrong because i keep getting the wrong answer for vf. It is kind of strange because my units in every part of solving the equation come out correctly.

Any suggestions?
 
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Are you manipulating the equality correctly? Perhaps if you were to show your steps we may be able to comment further.
 
Yes, sorry about that i should of.

First i converted km to meters because we work in the mks system

(4x10^5N)(2.5X10^6m)=w
=1x10^12J

1x10^12J=1/2(5x10^4kg)(vf^2)-1/2(5x10^4kg)(11,000m/s)^2

1x10^12J=1/2(5x10^4kg)(vf^2)-3.025J

from here i tried different ways to solve the equation but each way i still get the wrong answer. I added the 3.025J to both sides and then multiplied the 1/2(5x10^4). Then i divided the 4.025J by the 25,000kg and then solved for vf by taking the square root of both sides but for some reason my answer does not really come out relativlely close 9000m/s
 
rmarkatos said:
1x10^12J=1/2(5x10^4kg)(vf^2)-3.025J
Do you not think this number looks a little small?
 
Oh sorry a bit of typo there. It is supposed to be-3.025x10^12. Sorry i forgot to type that in there but when i was doing the problem i did have it correctly written.
 

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