Question

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

• A. 20 J at $x=2$ m
• B. 2 J at $x=20$ m
• C. 22 J at $x=2$ m
• D. 0 J at $x=2$ m

Hint:

This is in the vertex form of a parabola $y = k + (x-h)^2$. The vertex (minimum) is $(h, k)$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To find the minimum value of a function, we find the point where its first derivative is zero. Alternatively, observe the structure of the equation: a squared term $(x-2)^2$ is always $\ge 0$.
Step 2: Key Formula or Approach:
The minimum value occurs when the squared term is at its minimum, which is 0.
Step 3: Detailed Explanation:
The potential energy is $U(x) = 20 + (x-2)^2$. Since $(x-2)^2$ is a perfect square, its minimum possible value is $0$. This happens when $x - 2 = 0 \implies x = 2$ m. Substituting $x=2$ back into the original equation: \[ U_{min} = 20 + (2-2)^2 = 20 + 0 = 20 \text{ J} \]
Step 4: Final Answer:
The minimum potential energy is 20 J at position $x = 2$ m.