Question

the work done by the applied variable force ,F= x + x³ from x =0 to x= 2 where x is displacement , find work done Answer fast it's urgent

Answer 1

That is, the small (differential) amount of work done for a small (differential) displacement is

dW=F⃗ ⋅ds⃗  

So we can integrate and get the total work done as the integral of force over distance:

W=∫F⃗ ⋅ds⃗  

In the case that the force is constant over that displacement, we can take it out of the integral and integrate to get the form you stated as “known:”

W=∫F⃗ ⋅ds⃗ =F⃗ ⋅Δs⃗  

But if F is non-constant, as in your spring force F=kx, you can no longer pull it out of the integral. So the work done in that (1D) case is:

W=∫x2x1Fdx=∫x2x1kxdx=12k(x22−x21)

If you’ve gotten to energy then that (12kx2) might look familiar — it is the potential energy stored in a spring compressed/extended a distance x. Which it should be because potential energy is essentially just a convenient bookkeeping method for work.

Answer 2

Answer:

Hey mate...

That is, the small (differential) amount of work done for a small (differential) displacement is

dW=F⃗ ⋅ds⃗  

So we can integrate and get the total work done as the integral of force over distance:

W=∫F⃗ ⋅ds⃗  

In the case that the force is constant over that displacement, we can take it out of the integral and integrate to get the form you stated as “known:”

W=∫F⃗ ⋅ds⃗ =F⃗ ⋅Δs⃗  

But if F is non-constant, as in your spring force F=kx, you can no longer pull it out of the integral. So the work done in that (1D) case is:

W=∫x2x1Fdx=∫x2x1kxdx=12k(x22−x21)

If you’ve gotten to energy then that (12kx2) might look familiar — it is the potential energy stored in a spring compressed/extended a distance x. Which it should be because potential energy is essentially just a convenient bookkeeping method for work.

HOpe iT HelP YOu