How to work with non-constant forces?

  • Thread starter Thread starter PhizKid
  • Start date Start date
  • Tags Tags
    Forces Work
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 2K views
PhizKid
Messages
477
Reaction score
2

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.
 
Physics news on Phys.org
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.