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The potential energy of a particle varies the distance $x$ from a fixed origin as $U= \frac{A}{x^2} +B$ , where $A$ and $B$ are dimensional constants, then find the dimensional formula for $\frac{A^2}{B^2}.$

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Relate position, velocity, speed, and acceleration using derivatives

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Q. The potential energy of a particle varies the distance

x

from a fixed origin as

U= \frac{A}{x^2} +B

, where

A

and

B

are dimensional constants, then find the dimensional formula for

\frac{A^2}{B^2}.

Identify dimensions for U: Identify the dimensional formula for each term in the potential energy equation $U = Ax^2 + B$ . Potential energy ( $U$ ) has dimensions of energy, which is $ML^2T^{-2}$ .
Analyze term $Ax^2$ : Analyze the term $Ax^2$ . Since $x$ is a distance, it has dimensions of $L$ . Therefore, $x^2$ has dimensions of $L^2$ . To maintain dimensional consistency, $A$ must have dimensions of $M/T^2$ so that $Ax^2$ has dimensions of $ML^2T^{-2}$ .
Consider term B: Consider the term $B$ . Since $U = Ax^2 + B$ and both terms must have the same dimensions, $B$ must also have dimensions of $ML^2T^{-2}$ .
Calculate dimensions of $AB^2$ : Calculate the dimensions of $AB^2$ . Using the dimensions of $A (M/T^2)$ and $B (ML^2T^{-2})$ , $AB^2 = (M/T^2)(ML^2T^{-2})^2$ . Simplifying, $AB^2 = M^3L^6T^{-6}$ .

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