91Ó°ÊÓ

Given the graph which shows the variation of the conservative force F_X with x.

Force acting on the particle is conservative. So the work done by this conservative force is stored as the potential energy of the particle.

The work done by the conservative is calculated by measuring the area under the curve in the $F_x-x$graph. this work done is equal to the potential energy of the particle that position.

hence;

Potential energy = Area under $F_x-x$graph

potential energy at the following position is given by.

role="math" localid="1647781672256" $$\begin{gathered} Atx=1.0m \\ Potentialenergy=20N×1m=20Nm=20J \\ Atx=2.0m \\ Potentialenergy=20N×2m=40Nm=40J \\ Atx=3.0m \\ Potentialenergy=20N×3m=60Nm=60J \\ Atx=4.0m \\ Potentialenergy=\frac{1}{2}×20N×1m+20N×3m=70Nm=70J \end{gathered}$$

Given the graph which shows the variation of the conservative force F_X with x.

The turning point occurs where the total energy line crosses the potential energy curve.

Here, the potential energy at the,$x=1.0m$is found to be $20J$.

localid="1647783037125" $$\begin{gathered} Kineticenergy=\frac{1}{2}×m×v^2 \\ =\frac{1}{2}×100g×{\left(25m/s\right)}^2 \\ =\frac{1}{2}×\frac{100}{1000}kg×{\left(25m/s\right)}^2 \\ =31.25kg{m}^2/s^2 \\ =31.25Nm \\ =31.25J \end{gathered}$$

Total energy = Kinetic energy + potential energy

$$\begin{gathered} =31.25J+20J \\ =51.25J \end{gathered}$$

The graph can be drawn showing the total energy (TE) and the variation of potential energy.

The turning point occurs where the total energy line crosses the potential energy curve. We can see from the graph this is at approximately $2.5m$. For a more accurate value, the potential energy function is $U=20xJ$

The TE line crosses at the point where

$$\begin{gathered} 20x=51.25 \\ x=2.56m \end{gathered}$$