Archery!

Use The Law of Conservation of Energy to solve for the Elastic Potential Energy.

Conservation of Energy can be written as:

\begin{equation*}
E_i = E_f.
\end{equation*}

In this case:

\begin{equation*}
{U_{elast}+ mgh_i} = {mgh_f + \frac{1}{2}mv_f^2}.
\end{equation*}

Solving for \(U_{elast}\) and plugging in numerical values, we get:

\begin{equation*}
U_{elast}= 11.9 \, \text{J}.
\end{equation*}

To find the elastic energy of the bow we will use conservation of energy. Now, the mechanical energy of the arrow alone, or of the bow alone, is not conserved. The reason is that the bow is giving its energy to the arrow, so the arrow’s energy increases and the bow’s energy decreases. However, all the elastic energy of the bow is transmitted to the arrow, and so the energy gained by the arrow is exactly the energy lost by the bow. This means that the total mechanical energy of the bow-arrow system is indeed conserved, and so we can use conservation of energy to solve the problem (this illustrates that it is very important in energy problems to identify correctly the system whose conservation of energy we will study).

The total initial energy of the bow-arrow system is given by

\begin{equation}
\label{Arqueria_energiaElasticaInicial}
E_i = U_{elast} + U_{ig} + K_i,
\end{equation}

where \(U_{elast}\) is the elastic energy of the bow (that we are solving for), \(U_{ig}\) is the gravitational potential energy of the arrow, \( K_i \) is the initial kinetic energy of the arrow (the bow does not have gravitational potential energy, nor kinetic energy because we are neglecting its mass, and both of these types of energies depend on the mass). But initially the arrow is not moving, and so \(K_i\) is zero:

\begin{equation}
\label{Arqueria_energiaElasticaInicialSinCinetica}
E_i = U_{elast} + U_{ig}.
\end{equation}

Also, we know that \(U_{ig}=mgh_i\), where \(h_i\) is the initial height (that we know). This height is measured with respect to the floor, which means that it is convenient to use a coordinate system located on the floor (see figure 1).

Figure 1: We place the coordinate system on the ground at Ben’s position where the initial height is \(h_i\).

Then equation \eqref{Arqueria_energiaElasticaInicialSinCinetica} becomes:

\begin{equation}
E_f = U_{elast} + {mgh_i}.
\end{equation}

The final energy of the system is given by

\begin{equation}
\label{Arqueria_energiaElasticaFinal}
E_i = U_{fg} + K_f,
\end{equation}

where \(K_f\) is the final kinetic energy (not zero anymore) and \(U_{fg}\) is the final potential energy of the arrow (notice that there is no longer elastic energy because the bow is no longer tensed). Since the kinetic energy is given by \(1/2mv_f^2\) and \(U_{fg}\) by \(mgh_f\), we can write equation \eqref{Arqueria_energiaElasticaFinal} as

\begin{equation}
\label{Arqueria_energiaElasticaFinalConVariables}
E_f = {mgh_f} + {\frac{1}{2}mv_{f}^2}.
\end{equation}

Finally, we can use conservation of energy:

\begin{equation}
\label{Arqueria_conservacionEnergia}
{U_{elast}+ mgh_i} = {mgh_f + \frac{1}{2}mv_f^2}.
\end{equation}

If we move \(mgh_i\) to the right, we get

\begin{equation}
U_{elast} = mgh_f + \frac{1}{2}mv_f^2 – mgh_i.
\end{equation}

Finally, let’s insert the known values:

\begin{equation}
U_{elast} = {(0.03\, \text{kg})} {(9.8 \, \text{m/s}^2)} {(2\, \text{m})} + \frac{1}{2} {(0.03\, \text{kg})} ({28 \, \text{m/s}})^2 – {(0.03\, \text{kg})} {(9.8 \, \text{m/s}^2)} {(1.5\, \text{m})}.
\end{equation}

The result is

\begin{equation}
U_{elast} = 11.9 \, \text{J}.
\end{equation}

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