Question

A system in which only one particle can move has the potential energy shown in $FIGUREEX10.30$. What is the x-component of the force on the particle at $x=5,15,$and $25cm$?

Short answer

x-component of force at$x=5cm$=$100N$.

x-component of force at$x=15cm$=$0N$.

x-component of force at$x=25cm$=$-50N$.

Step-by-step solution

Force : 

The force exerted on a mass item that causes it to alter velocity.

Explanation : 

The equation gives the x-component of force on the potential energy diagram.

$F_x=-\frac{dU}{dx}$

Particle's potential energy diagram,

The slope of the potential energy diagram is the x-component of force.

At $x=5cm$, the x-component of force(slope) is,

$$\begin{gathered} F_{x=15}=-\frac{\left(10-0\right)J}{\left(0-10\right)cm} \\ =1\left(J/cm\right) \\ =100N\left(1J/cm=\frac{1N-m}{0.01m}=100N\right) \end{gathered}$$

At $x=15cm$,the x-component of force(slope) is,

$$\begin{gathered} F_{x=15}=-\frac{\left(0-0\right)J}{\left(10-20\right)cm} \\ =0\left(J/cm\right) \\ =0N\left(1J/cm=\frac{1N-m}{0.01m}=100N\right) \end{gathered}$$

At$x=25cm$,the x-component of force(slope) is,

$$\begin{gathered} F_{x=15}=-\frac{\left(0-10\right)J}{\left(20-40\right)cm} \\ =-0.5\left(J/cm\right) \\ =-50N\left(0.5J/cm=\frac{0.5N-m}{0.01m}=50N\right) \end{gathered}$$