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# the potential energy between two idential point like objects of the same mass,mi is given by the relation U(r)=mA[(r0/r)^12-2(ro/r)^6]?

here r is the distance between the objects r0 is the equilibrium distance where the net force on objects is zero and A is a constant.

What is unit of A?

What is the minimum value of the potential energy?

What is the magn. of the force applied by one of the objects on the other at the distance that the potential energy becomes minimum ?

WHat is the magnitude of the force applied on each objects as a function of the distance r?

Consider that one of the objects is fixed. What is the min work that must be done to bring the other object from a distance r0 to 2r0?

### 2 Answers

- NCSLv 74 years agoFavorite Answer
U(r) = mA[(r0/r)^12 - 2(r0/r)^6]

(i) units of A must be m²/s² assuming m is the mass in kg and U is in Joules.

(ii) dU/dr = 0 = mA[-12*r0^12 / r^13 + 2*6*r0^6 / r^7]

and this is zero where

12*r0^12 / r^13 = 12*r0^6 / r^7

r0^6 / r^6 = 1

r = r0

so

U(r0) = mA[(r0/r0)^12 - 2(r0/r0)^6] = mA(1 - 2) = -mA ◄ min PE

(iii) F = dU/dr = 0 at r = r0 ◄

(iv) F(r) = mA[-12*r0^12 / r^13 + 2*6*r0^6 / r^7] = 12mA[*r0^6 / r^7 - r0^12 / r^13]

(v) We know U(r0) = -mA

U(2*r0) = mA*[(r0 / 2*r0)^12 - 2*(r0 / 2*r0)^6] = mA[½^12 - 2*½^6]

U(2*r0) = -0.031mA

so we must do (-0.031 - -1)mA = 0.969mA ◄ required work

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