How to work with non-constant forces?

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The discussion revolves around calculating the speed of a 1 kg block pushed by a non-constant force defined as x^2 at a displacement of 10 meters. The initial approach involved using Newton's second law, leading to an incorrect velocity calculation of 333.3 m/s. The correct method involves applying the work-energy theorem, integrating the force to find the total work done, and equating it to the change in kinetic energy. After integrating, the total work done is found to be (x^3)/3, leading to a final speed of approximately 25.8 m/s at 10 meters. The integration approach is confirmed as the appropriate method for dealing with non-constant forces.
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Homework Statement


A 1 kg. block at rest is pushed with a force of x^2, where x is the displacement (in meters). What is the speed of the object at 10 meters?


Homework Equations


F = ma


The Attempt at a Solution


I did x^2 = ma and found the acceleration by dividing the mass which is just 1. So a = x^2. I integrated to find the velocity function so I got [(x^3)/3] + C. It's initially at rest so at x = 0, the velocity is 0 which means C = 0 so I can get rid of that. If I plug in 10 meters into the velocity equation: (10^3)/3 I get 333.3 m/s. Solution says this is wrong but I have never worked with integration before so I'm not sure how to do this.
 
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I would consider using the Work energy theorem.
 
So the total work done is F*d. So that's 10x^2. Initial velocity is 0 at rest so change in KE = Work done: (1/2)mv^2 = 10x^2. Solve for v and it's sqrt(20)*x. At x = 10, I get v = 10sqrt(20). Is this right?
 
PhizKid said:
So the total work done is F*d.
The force is not constant. You'll have to integrate: W = ∫F(x)dx
 
So the total work done is (x^3)/3 ? So can I do:

(x^3)/3 = (1/2)mv^2
666.66 = v^2
v = 25.8 m/s ?
 
PhizKid said:
So the total work done is (x^3)/3 ? So can I do:

(x^3)/3 = (1/2)mv^2
666.66 = v^2
v = 25.8 m/s ?
Looks good to me.
 
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