Answer
a) Left.
b) Yes.
c) $200\;\rm N/m$
d) $19\;\rm m/s$
Work Step by Step
We can treat the rubber band and the rock as an isolated system. The force exerted by the rubber band is horizontal as we see in the given graph which means that there are no external forces exerted on our system and hence its energy is conserved.
a) The given graph is shown the force exerted by the rubber band on the rock.
$$\sum F_{(x, \rm \;on\;rock)}=F_x=-kx$$
and we can see that the rock moves to the left since $x$ is negative in the given figure.
$$ F_x=-k(-x)=kx$$
Thus, the force exerted by the rubber band on the rock is toward the right, and hence the rubber was stretched to the left.
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b) We know that Hooke's law is given by $F_{sp}=-kx$ which means that the elastic force is a function in the stretched (or compressed) distance that gave us a straight line that has a slope of $k$.
And we can see that the given data is a straight line for $F_x$ as a function in $x$.
So, yes, it obeys Hooke's law.
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c) The rubber band’s spring constant is given by the slope of the straight line.
$$k={\rm slope}=\dfrac{\Delta F_x}{\Delta x}=\dfrac{60}{0.30}=\color{red}{\bf200}\;\rm N/m$$
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d) Since the energy is conserved, so the elastic potential energy will be fully converted to kinetic energy. So that
$$U_i+\overbrace{K_i}^{0}=\overbrace{U_f}^{0}+K_f$$
$$\frac{1}{2}kx^2=\frac{1}{2}mv_f^2$$
Solving for $v$;
$$ v_f =\sqrt{\dfrac{kx^2}{m} }=\sqrt{\dfrac{k }{m} }\;x$$
Plugging the known;
$$ v_f =\sqrt{\dfrac{200 }{0.05} }\times 0.3=\color{red}{\bf 19}\;\rm m/s$$
