The variation of potential energy U of a body moving along x-axis varies with its position (x) as shown in figure The body is in equilibrium state at

Question

The variation of potential energy U of a body moving along x-axis varies with its position (x) as shown in figure <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/1df7ff31d29f6836247ccca673323747-.png" /> The body is in equilibrium state at (A) A (B) B (C) C (D) Both A C

Tardigrade Admin · Accepted Answer

at B, (d U/d x)=0 (Slope of U-x curve) ⇒ F=0 at B, So its a position of equilibrium

Q. The variation of potential energy $U$ of a body moving along $x$ -axis varies with its position $(x)$ as shown in figure

The body is in equilibrium state at

A

B

C

Both A & C

at $B, \frac{d U}{d x} = 0$ (Slope of $U - x$ curve)
$\Rightarrow F = 0$ at $B$ , So its a position of equilibrium

Q. The variation of potential energy U of a body moving along x-axis varies with its position (x) as shown in figure The body is in equilibrium state at

A

B

C

Both A & C

at B,dxdU​=0 (Slope of U−x curve) ⇒F=0 at B, So its a position of equilibrium

Q. The variation of potential energy $U$ of a body moving along $x$ -axis varies with its position $(x)$ as shown in figure

The body is in equilibrium state at

at $B, \frac{d U}{d x} = 0$ (Slope of $U - x$ curve)
$\Rightarrow F = 0$ at $B$ , So its a position of equilibrium