Calculating Work Done on a Particle Using Basic Formulas

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Homework Help Overview

The discussion revolves around calculating the work done on a particle-like object under the influence of a variable force, with the position described by a cubic function of time. Participants are exploring the implications of the given position function and the relationship between position, force, and work.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial attempts to calculate work using distance and acceleration, but express uncertainty about the correctness of their methods. Questions arise regarding the application of integral formulas for work and how to handle variable forces.

Discussion Status

The conversation is ongoing, with participants seeking clarification on the relationship between position, mass, and force. Some guidance has been offered regarding the need to express the force acting on the particle, and there is an invitation for the original poster to share their equations for further assistance.

Contextual Notes

There is a mention of the original poster's confusion regarding the integration of position and mass to find work, indicating a potential gap in understanding the application of work formulas in the context of variable forces.

Jrlinton
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Homework Statement


A single force acts on a 3.3 kg particle-like object in such a way that the position of the object as a function of time is given by x = 2.1t - 1.4t2 + 1.7t3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 4.8 s.

Homework Equations


Basic work formulas

The Attempt at a Solution


So i found the distance of the movement by plugging 4.8 for t and got 165.83 m, found the second derivative of the position formula to get the acceleration function, got 46.16m/s/s at 4.8s. I multiplied these quantities as well as the mass to get 25260.6 J. This was obviously wrong
 
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Jrlinton said:

Homework Statement


A single force acts on a 3.3 kg particle-like object in such a way that the position of the object as a function of time is given by x = 2.1t - 1.4t2 + 1.7t3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 4.8 s.

Homework Equations


Basic work formulas

The Attempt at a Solution


So i found the distance of the movement by plugging 4.8 for t and got 165.83 m, found the second derivative of the position formula to get the acceleration function, got 46.16m/s/s at 4.8s. I multiplied these quantities as well as the mass to get 25260.6 J. This was obviously wrong
Are you familiar with the integral formula for calculating the work?
 
Yes but I am ignorant as to how that would work with the givens being position and mass
 
You have a variable force. How do you find work done if you have a variable force acting on a particle?

EDIT: First, what's the expression for the force acting on the particle?
 
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Jrlinton said:
Yes but I am ignorant as to how that would work with the givens being position and mass
Well then please show them to us and start working with the equations. That will help us a lot in helping you. If you can use LaTeX in your posts of the equations, that would be best. I'll find the link to the PF tutorial on LaTeX equations in a sec...
 

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