Answer
See the detailed answer below.
Work Step by Step
From the given graph is it obvious that the particle undergoes three stages.
- First, from $x=0$ m to $x=3$ m, a constant force of 20 N.
- Second, from $x=3$ m to $x=4$ m, the force decreases unit it became zero.
- Third, from $x=4$ m to $x=5$ m, the force is zero.
_________________________________________
a) We need to draw the graph of potential energy $U$ versus displacement $x$ by using the given graph of force $F$ versus $x$.
We know that the potential energy is given by
$$dU=-dF_xdx $$
which is the area under the force versus displacement curve, as our given graph.
Thus,
$$ \Delta U=-\text{ Area under the given curve}=A$$
Noting that $1,2$ and $3$ are for the three stages as we mentioned above.
We are given that at $x=0$, $U=0$, so that
$$\Delta U_1=-A_1=-( 3\times -20 )=\bf 60\;\rm J$$
Thus, at $x=0$ m, $U=0$ J, and at $x=3$ m, $U=60$ J.
Hence, this stage is a straight line.
$$\Delta U_2=-A_2=- \left(\frac{1}{2}\times 1\times -20\right)=\bf 10\;\rm J$$
Thus, at $x=4$, $U=\Delta U_1+\Delta U_2=60+10=\bf 70 \;\rm J$
This stage is a a quadratic dependence on $x$ since $F$ changes linearly.
$$\Delta U_2=-A_3 =\bf 0\;\rm J$$
See the figure below.
_________________________________________
b) We know that the mechanical energy of the particle is given by
$$E_{mech}=K+U$$
And we know the potential energy of the particle at $x=1$, from the figure below. We also know its speed at the same point, so we can find its mechanical energy.
$$E_{mech}=\frac{1}{2}mv^2+U=\frac{1}{2}(0.1)(25^2)+20$$
$$E_{mech}=\color{red}{\bf 51.25}\;\rm J$$
_________________________________________
c) See the blue line in the figure below.
_________________________________________
d) The turning point occurs at $v=0$, which means it occurs when the total energy of the particle is now a potential energy.
$$U=51.15\;\rm J$$
which is at $$\color{red}{\bf x=2.56}\;\rm m$$
