Work and energy -- Change in KE due to a force F acting on a mass

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Discussion Overview

The discussion revolves around the application of work and energy principles, specifically focusing on the change in kinetic energy due to a force acting on a mass. Participants are addressing a problem statement involving a force vector and its implications for motion, integration, and initial conditions.

Discussion Character

  • Homework-related
  • Technical explanation

Main Points Raised

  • One participant points out that the force vector provided by @Max2020 is part of the problem statement and not relevant for the integration process.
  • Another participant emphasizes the importance of the equation ##\vec F = m\vec a## and the work done equation ##W = \int \vec F\cdot\vec s## as relevant to the discussion.
  • There is a mention of an initial velocity vector ##v_0 = \hat \jmath + 2\,\hat k## that seems to be missing from @Max2020's work, raising questions about the completeness of the provided information.
  • Participants express confusion regarding the format of the posted image, which is rotated, potentially complicating the understanding of the problem.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on how to proceed with the problem due to the missing initial conditions and the format of the image. Multiple viewpoints regarding the relevance of the equations and the integration process remain present.

Contextual Notes

There are limitations regarding the clarity of the problem due to the rotated image and the absence of certain initial conditions, which may affect the ability to fully address the question posed by @Max2020.

Max2020
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Homework Statement
a force of F = 3i + (6t ^ 2) j - tk acts on a particle of mass 2 kg, where F is given in newtons and t in seconds. if the initial velocity of the particle is Vo = j + 2k, in meters per second. () What is the work done by force F during the interval 0 <= t <= 2? () Using the definition of kinetic energy (K = mv ^ 2/2), find the kinetic energy variation ∆K of the particle in the same range.
Relevant Equations
F= 3i +(6t^2) j -tk
20210426_150244.jpg
 
Last edited by a moderator:
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Hello @Max2020 ,
:welcome: ##\qquad !##​

Did you notice your picture is rotated by 90 degrees ? It hurts to look at it !

Your $$\vec F= 3\,\hat\imath +6t^2 \,\hat\jmath -t\,\hat k $$ is not a relevant equation: it is part of the problem statement. Fortunately I do find a relevant equation (##\vec F = m\vec a##) in your picture if I almost break my neck.

The other relevant equation is ##W = \int \vec F\cdot\vec s##, also to be found in your work.

So you try to integrate ##\vec F = m\vec a\ ## twice. But I miss the given ##v_0 = \hat \jmath + 2\,\hat k\ ## there ?

Or do you have some other question that I somehow missed ?

##\ ##
 
BvU said:
Hello @Max2020 ,
:welcome: ##\qquad !##​

Did you notice your picture is rotated by 90 degrees ? It hurts to look at it !

Your $$\vec F= 3\,\hat\imath +6t^2 \,\hat\jmath -t\,\hat k $$ is not a relevant equation: it is part of the problem statement. Fortunately I do find a relevant equation (##\vec F = m\vec a##) in your picture if I almost break my neck.

The other relevant equation is ##W = \int \vec F\cdot\vec s##, also to be found in your work.

So you try to integrate ##\vec F = m\vec a\ ## twice. But I miss the given ##v_0 = \hat \jmath + 2\,\hat k\ ## there ?

Or do you have some other question that I somehow missed ?

##\ ##
I'm sorry for posting the photo like this. I still can't see a way to resolve this issue.
 
Max2020 said:
I'm sorry for posting the photo like this. I still can't see a way to resolve this issue.
1619481818607.png


:smile:
 

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