Work and Kinetic Energy: Calculating Work and Speed with Varying Force

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Homework Help Overview

The problem involves calculating work done by a varying force on a mass and determining the speed of the mass at a specific displacement. The subject area is work and kinetic energy in the context of classical mechanics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of work using the given force equation and question the method of applying the force to find work. There is a focus on the distinction between calculating force and work, with some participants suggesting the use of integration for non-constant forces.

Discussion Status

The discussion is active, with participants exploring different interpretations of how to calculate work done by the varying force. Some guidance has been offered regarding the need for integration, indicating a productive direction in the conversation.

Contextual Notes

Participants are working under the assumption that no other forces are acting on the mass, and there is a specific displacement at which the calculations are to be made. There is also a noted confusion regarding the application of force to calculate work.

shenwei1988
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The magnitude of a certain one-dimensional force varies according to:

F = 6.68x^2 + 1.56

where x is the displacement from the origin in meters, and F is the force in Newtons. At t = 0, a 677 g mass is at the origin moving in the positive x-direction at speed 8.46 m/s when this force begins to act on it.
ASSUME: there are no other forces acting on the mass.

a) How much work is done by the force on the mass when it reaches x = 2.78 m?
b) What is the speed of the mass when it reaches 2.78 m?




W=FS


i put x=2.78 into the equation. w=(6.68*2.78^2)*2.78
and get the wrong answer.







The Attempt at a Solution

 
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shenwei1988 said:
i put x=2.78 into the equation. w=(6.68*2.78^2)*2.78
and get the wrong answer.

If you put the value of x, you get the force, not the work. Think about what is work done in terms of force and distance.
 
not, i put the value of x. and get the force, then use the force * distance.w=(6.68*2.78^2+1.56)*2.78
 
When the force is not constant, then the work done is Integral(from x1 to x2)[Fdx].
 

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