Answer
See the detailed answer below.
Work Step by Step
a) We know that the force is given by
$$F=\dfrac{-dU}{dx}$$
which is the slope of the given graph.
$$F=-{\rm Slope}$$
We have 5 stages of potential energy in the given graph which are
- The first stage from $x=0$ m to $x=0.01$ m
$$F_1=-{\rm Slope}_1=-\dfrac{\Delta U}{\Delta x}=\dfrac{0-4}{0.01-0}=\bf 400\;\rm N$$
- The second stage from $x=0.01$ m to $x=0.03$ m
$$F_2=-{\rm Slope}_2=-\dfrac{\Delta U}{\Delta x}=\dfrac{0-0}{0.03-1}=\bf 0\;\rm N$$
- The third stage from $x=0.03$ m to $x=0.05$ m
$$F_3=-{\rm Slope}_3 =-\dfrac{2-0}{0.05-0.03}=\bf -100\;\rm N$$
- The fourth stage from $x=0.05$ m to $x=0.07$ m
$$F_4=-{\rm Slope}_4 =-\dfrac{2-2}{0.07-0.05 }=\bf 0\;\rm N$$
- The fifth stage from $x=0.07$ m to $x=0.08$ m
$$F_5=-{\rm Slope}_5 =-\dfrac{6-2}{0.08-0.07 }=\bf -400\;\rm N$$
Now we can easily draw the force versus displacement graph, as you see below.
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b) As shown below, the force from 2 cm to 3 cm is zero and it is also zero from 5 cm to 6 cm. So, we actually need to find the work done by the force from 3 cm to 5 cm (which is the force of the third stage, as we see above in part [a]).
$$W=W_{2\rightarrow3 }+W_{3\rightarrow5}+W_{5\rightarrow6 }$$
$$W=0+(-100\times [0.05-0.03])+0$$
$$W=\color{red}{\bf -2}\;\rm J$$
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c)
At $x=2$ cm, we know from the given figure that the particle's potential energy is $U_2=0$ J and its kinetic energy is $K_2$.
At $x=6$ cm, we know from the given figure that the particle's potential energy is $U_6=2$ J and its kinetic energy is $K_6=\frac{1}{2}mv_f^2=\frac{1}{2}\times (10\times 10^{-3})\times (10^2)=\bf 0.5$ J.
Using the conservation of energy;
$$K_2+U_2=K_6+U_6$$
Plugging the known;
$$\frac{1}{2}mv_i^2+0=0.5+2$$
Thus,
$$v_i^2=\dfrac{2.5\times 2}{m}$$
$$v_i =\sqrt{\dfrac{2.5\times 2}{10\times 10^{-3}}}$$
$$v_i=\color{red}{\bf 22.4}\;\rm m/s$$
