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

(a)

 

In this problem, F(x) = F_x(x) = -Ax + Bx⁶

 

U(x) = -∫F(x)dx = -∫(-Ax + Bx⁶)dx = Ax²/2 - Bx⁷/7 + C

 

U(0) = 0 ⇒ C = 0

 

U(x) = Ax²/2 - Bx⁷/7

 

(b)

 

ΔU = U₂ - U₁ = U(x₂) - U(x₁) = U(3.80 m) - U(1.60 m)

 

Just evaluate U(x) at the two end points 1.60 m and 3.80 m, and take the difference in order U₂ - U₁. 

 

From energy conservation, which applies in this problem,

 

ΔK = -ΔU