Question

1)A single conservative force acting on a particle within a system varies as 𝐹⃗ = (−𝐴𝑥 + 𝐵𝑥

2

)𝑖, ̂

where A and B are constants, 𝐹⃗ is in newtons, and x is in meters. (a) Calculate the potential 

energy function U(x) associated with this force for the system, taking U=0 at x=0. Find (b)

the change in potential energy and (c) the change in kinetic energy of the system as the 

particle moves from x=2m to x-3m.

Answer 1

Expression of conservative force, $\vec F=(-Ax+Bx^2)\hat i\space N$

(a) Potential energy,

$d U=\vec F.dx$

$d U=(-Ax+Bx^2)dx$

At x = 0, U = 0 (given)

$\int_0^U d U=\int_0^x(-Ax+Bx^2)dx$

$U(x)=\dfrac{-Ax^2}{2}+\dfrac{Bx^3}{3}$

(b) Change in potential energy as x changes from 2 to 3

$U(3)-U(2)=\bigg|\dfrac{-Ax^2}{2}+\dfrac{Bx^3}{3}\bigg|_2^3$

$\Delta U=\dfrac{-9A}{2}+\dfrac{27B}{3}+\dfrac{4A}{2}-\dfrac{8B}{3}$

$\Delta U=\dfrac{-5A}{2}+\dfrac{19B}{3}$

(c)  If we consider the particle alone as a system, the change in its kinetic energy is the work done by the force on the particle:

$W=ΔK$

For the entire system of which this particle is a member, this work is internal work and equal to the negative of the change in potential energy of the system:

$ΔK=−ΔU=2.5A-6.33B$