A force on an object is given by F(x)= ( -4.00 N/m)x + ( 2.00 N/m³)x³.What is the change in potential energy in moving from x = 1.00 m to x = 2.00 m? A)10.0 J B)-1.50 J C)-10.0 J D)1.50 J E)12.0 J

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Quizplus · Accepted Answer

The change in potential energy ($\Delta U$) when moving from $x_1$ to $x_2$ can be found by integrating the force $F(x)$ with respect to $x$ over the interval from $x_1$ to $x_2$. The given force is $F(x) = (-4.00 \, \text{N/m})x + (2.00 \, \text{N/m}^3)x^3$. The integral of this force from $x = 1.00 \, \text{m}$ to $x = 2.00 \, \text{m}$ gives the change in potential energy. The integral of $F(x)$ with respect to $x$ is $-2.00x^2 + (0.5)x^4$, evaluated from 1 to 2. This results in $-2.00(2^2) + 0.5(2^4) - (-2.00(1^2) + 0.5(1^4)) = -8 + 8 - (-2 + 0.5) = -1.5 \, \text{J}$. Therefore, the change in potential energy is $-1.50 \, \text{J}$.

A Force on an Object Is Given by F(x)=