Answer to Question #197898 in Mechanics | Relativity for DANI

Question

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

Gives

$F=(-Ax+Bx^2)i$

$U=-\intop F.dr$

Put Value

$$U=-\smallint(-Ax+Bx^2)i.(dxi+dyj+dzk)$$

$U(x)=\smallint(Ax-Bx^2)dx$

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

Then

x=0,U=0

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

$U(2)=\frac{A2^2}{2}-\frac{B2^3}{3}$

$U(2)=2A-2.67B$

U(3)=$\frac{A3^2}{2}-\frac{B3^3}{3}$

U(3)=4.5A-9B

∆U=4.5A -9B -(2A-2.67B)=2.5A -6.33B

∆U=2.5A-6.33B

If consider particle alone as a system

The change in its kinetic energy is the work done by the force

On the particle

W=∆K

For the entire system of which this particle of a mamber

This work is internal work and equal to the negative of the change in potential energy of the system

∆K=-∆U= -2.5A+6.33B