Answer
a) $\rm \dfrac{N}{m^3}=\rm\dfrac{kg}{s^2.m^2}$
b) $U= \frac{1}{4}qx^4$
c) $10\;\rm m/s$
Work Step by Step
a) We need to solve the given formula for $q$ so we can find its units.
$$F_x=-qx^3$$
Hence,
$$q=\dfrac{-F_x}{x^3}$$
where the force is measured in Newtons and the distance is measured in meters.
Thus, the units of $q$ are given by
$$\rm\boxed{\rm \dfrac{N}{m^3}}=\dfrac{kg.m}{s^2.m^3}=\boxed{\rm\dfrac{kg}{s^2.m^2}}$$
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b) To find a formula for the potential energy of the sprong, we need to recall that
$$F_x=\dfrac{-dU}{dx}$$
Thus,
$$dU=-F_xdx$$
$$\int_0^UdU=\int _{x_e}^x-F_xdx$$
Hence,
$$ U=-\int _{x_e}^xF_xdx=-\int _{x_e}^x(-qx^3)dx$$
$$ U=q \int _{x_e}^x x^3dx=q\dfrac{x^4}{4}\bigg| _{x_e}^x$$
$$ U= \dfrac{q}{4}\left[x^4-x_e^4\right] $$
where $x_e=0$, so that
$$ \boxed{U= \frac{1}{4}qx^4}$$
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c) Since the barrel is frictionless, then the energy is conserved.
This means that the elastic potential energy stored in the sprong will be completely converted to a kinetic energy.
Thus,
$$U_i=K_f$$
Plugging from the boxed formula above.
$$\frac{1}{4}qx^4=\frac{1}{2}mv^2$$
Solving for $v$;
$$v=\sqrt{\dfrac{\frac{1}{2}qx^4}{m}}=\sqrt{\dfrac{\frac{1}{2}q }{m}} x^2$$
Plugging the known;
$$v=\sqrt{\dfrac{\frac{1}{2}(40000) }{20\times 10^{-3}}} \times 0.10^2=\color{red}{\bf 10}\;\rm m/s$$
