(II) An 85-g arrow is fired from a bow whose string exerts an average force of 105 N on the arrow over a distance of 75 cm. What is the speed of the arrow as it leaves the bow?

The mass of the arrow is\(m = 85\;{\rm{g}} = 0.085\;{\rm{kg}}\).

The initial speed of the arrow is\({v_{\rm{i}}} = 0\;{\rm{m/s}}\).

The arrow covers a distance of\(d = 75\;{\rm{cm}} = 0.75\;{\rm{m}}\).

The force applied by the bow on the arrow is \(F = 105\;{\rm{N}}\).

The work done on the arrow is equal to the kinetic energy of the bow when it just leaves the bow.

The initial kinetic energy of the arrow is zero as it starts from rest.

Let the final speed of the arrow be\({v_{\rm{f}}}\).Then, the final kinetic energy of the base arrow becomes \(\frac{1}{2}mv_{\rm{f}}^2\).

The change in the kinetic energy of the arrow is:

\(\begin{array}{c}\Delta K = \frac{1}{2}mv_{\rm{f}}^2 - 0\\ = \frac{1}{2}mv_{\rm{f}}^2\end{array}\)

Now, the work done by the force is \(W = Fd\).

Now, from the work-energy theorem, you can write:

\(\begin{array}{c}\Delta K = W\\\frac{1}{2}mv_{\rm{f}}^2 = Fd\\v_{\rm{f}}^2 = \frac{{2Fd}}{m}\\{v_{\rm{f}}} = \sqrt {\frac{{2Fd}}{m}} \end{array}\)

Now, substituting the values in the above equation, you will get:

\(\begin{array}{c}{v_{\rm{f}}} = \sqrt {\frac{{2 \times 105\;{\rm{N}} \times 0.75\;{\rm{m}}}}{{0.085\;{\rm{kg}}}}} \\ = 43.05\;{\rm{m/s}}\\ \approx {\rm{43 m/s}}\end{array}\)

Hence, the speed of the arrow is \(43\;{\rm{m/s}}\) when it leaves the bow.