Question

The potential energy of a particle is given by \(U(x)=20+(x-2)^2\), where \(U\) is in joules and \(x\) is in meters. The minimum potential energy and the position where it occurs are

• A. \(20\ \text{J at }x=2\)
• B. \(2\ \text{J at }x=20\ \text{m}\)
• C. \(22\ \text{J at }x=2\ \text{m}\)
• D. \(0\ \text{J at }x=2\ \text{m}\)

Hint:

For expressions like \(a+(x-b)^2\), the minimum value is \(a\), occurring at \(x=b\).

Answer 1

The Correct Option is A

Concept: A square term is always non-negative: \[ (x-2)^2\geq 0 \] So the minimum value occurs when the square term becomes zero.

Step 1: Given: \[ U(x)=20+(x-2)^2 \]

Step 2: Since: \[ (x-2)^2\geq 0 \] The minimum value of \((x-2)^2\) is: \[ 0 \]

Step 3: This happens when: \[ x-2=0 \] \[ x=2 \]

Step 4: Substitute \(x=2\) into \(U(x)\). \[ U(2)=20+(2-2)^2 \] \[ U(2)=20+0 \] \[ U(2)=20\ \text{J} \] Therefore, minimum potential energy is: \[ \boxed{20\ \text{J at }x=2} \]