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

Step 1: Understanding the Question:
Potential energy \(U(x)\) is minimum where its derivative with respect to \(x\) is zero (equilibrium point) and the second derivative is positive.
Mathematically, we are looking for the minimum of a quadratic function.
Step 2: Key Formula or Approach:
To find the minimum, set \(\frac{dU}{dx} = 0\).
Substitute the resulting value of \(x\) back into \(U(x)\) to find the minimum energy value.
Step 3: Detailed Explanation:

Function: \(U(x) = 20 + (x - 2)^2\).

Differentiate with respect to \(x\):
\[ \frac{dU}{dx} = \frac{d}{dx}[20] + \frac{d}{dx}[(x - 2)^2] \]
\[ \frac{dU}{dx} = 0 + 2(x - 2) \cdot (1) = 2(x - 2) \]

Set the derivative to zero to find the critical point:
\[ 2(x - 2) = 0 \implies x = 2 \]

Since the expression involves a squared term \((x-2)^2\), it is always non-negative. Its minimum possible value is 0, which occurs when \(x = 2\).

Substituting \(x = 2\) into the equation for \(U(x)\):
\[ U_{\text{min}} = 20 + (2 - 2)^2 = 20 + 0^2 = 20 \text{ J} \]

Therefore, minimum potential energy is 20 J and it occurs at position \(x = 2\) m.

Step 4: Final Answer:
The particle has a minimum potential energy of 20 J at the equilibrium position \(x = 2\) m.