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

AI Thread Summary
The discussion focuses on calculating work done by a variable force defined by the equation F = 6.68x^2 + 1.56, acting on a 677 g mass initially moving at 8.46 m/s. A participant incorrectly attempts to calculate work by directly substituting displacement into the force equation. Another contributor clarifies that work should be calculated using the integral of force over distance, emphasizing that work done is not simply force multiplied by distance when the force varies. The correct approach involves integrating the force function from the initial to the final displacement. Understanding this concept is crucial for accurately determining both work and the resulting speed of the mass.
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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