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A particle is placed at the origin and a...

A particle is placed at the origin and a force F=Kx is acting on it (where k is a positive constant). If `U_((0))=0`, the graph of `U (x)` verses x will be (where U is the potential energy function.)

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To solve the problem, we need to find the potential energy function \( U(x) \) given the force \( F = kx \), where \( k \) is a positive constant. We know that the force is related to the potential energy by the equation: \[ F = -\frac{dU}{dx} \] ### Step 1: Relate Force to Potential Energy Given the force \( F = kx \), we can substitute this into the equation: \[ kx = -\frac{dU}{dx} \] ### Step 2: Rearranging the Equation Rearranging the equation gives us: \[ \frac{dU}{dx} = -kx \] ### Step 3: Integrate to Find Potential Energy Now, we need to integrate both sides with respect to \( x \): \[ dU = -kx \, dx \] Integrating both sides: \[ U = -\frac{1}{2} kx^2 + C \] ### Step 4: Determine the Constant of Integration We are given that \( U(0) = 0 \). Plugging in \( x = 0 \): \[ U(0) = -\frac{1}{2} k(0)^2 + C = 0 \implies C = 0 \] Thus, the potential energy function simplifies to: \[ U(x) = -\frac{1}{2} kx^2 \] ### Step 5: Analyze the Graph of \( U(x) \) The equation \( U(x) = -\frac{1}{2} kx^2 \) represents a downward-opening parabola. The vertex of this parabola is at the origin (0,0), and as \( x \) moves away from the origin in either direction, \( U(x) \) becomes more negative. ### Conclusion The graph of \( U(x) \) versus \( x \) is a downward-opening parabola symmetrical about the \( U \)-axis with the vertex at the origin.
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