PHYSICS QUESTION

Question

A single conservative force acting on a particle varies as vector F = (-Ax + Bx6)i hat N, where A and B are constants and x is in meters. Accurately round coefficients to three significant figures.

 

 

(a) Calculate the potential energy function U(x) associated with this force, taking U = 0 at x = 0. (Use A, B, and x as appropriate.)

 

U =

 

(b) Find the change in potential energy and change in kinetic energy as the particle moves from x = 1.60 m to x = 3.80 m. (Use A, B, and x as appropriate.)

 

ΔU =

 

ΔK =

 

Answer 1

Hi Porcia!

 

(a) The relationship between potential energy and force (in one dimension, as we have here) is that:

 

U(x) = -∫Fdx

 

In particular, in this case:

 

U(x) = -∫₀^x (-Ax+Bx⁶)dx

 

U(x) = - [-(A/2)x² + (B/7)x⁷]₀^x = (A/2)x² - (B/7)x⁷ - 0

 

(b)  We can now evaluate U(x) at x = 1.6 and x = 3.8, and thus calculate ΔU between those two points (because this is a conservative forces, the change will only depend on the endpoints, not the path taken from 1.6 m to 3.8 m).

 

ΔU = U_final - U_initial = U(3.8) - U(1.6) = (A/2)(3.8² - 1.6²) - (B/7)(3.8⁷ - 1.6⁷) = 11.88(A/2) - 11414(B/7)

 

If only the conservative force is acting, then the total change in mechanical energy must be zero over all intervals, meaning that whatever change we just calculated for potential energy, the kinetic energy change is just NEGATIVE that.  So:

 

ΔKE = 11414(B/7)-11.88(A/2)

 

I hope this helps!  If you want to go into any more detail or have any questions, just let me know