A single conservative force acting on a particle varies as `F=(-Ax+Bx^2)hatiN`, where A and B are constants and x is in meters. (a) Calculate the potential-energy function `U(x)` associated with this force, taking `U=0` at `x=0`.
(b) Find the change in potential energy and the change in kinetic energy as the particle moves from `x=2.00m` to `x=3.00m`.

Explanation

a. `U=-underset0oversetx int(-Ax+Bx^2)dx=(Ax^2)/(2)-(Bx^3)/(3)`
b. `DeltaU=DeltaU_(x=3m)-DeltaU_(x=2m)`
`=(A[(3.00^2)-(2.00)^2])/(2)-B([(3.00)^3-(2.00)^3])/(3)`
`=5.00/2A-19.0/3B`
`DeltaK=-DeltaU=(-5.00/2A+19.0/3B)`