Question

An archer draws her bow and stores 34.8 J of elastic potential energy in the bow. She releases the 63 g arrow, giving it an initial speed of 28 m/s. How efficient is the transfer of the elastic potential energy of the bow to the kinetic energy of the arrow?

Answer 1

Answer:

Approximately $71\%$.

Explanation:

The formula for the kinetic energy $\rm KE$ of an object is:

$\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2$,

where

• $m$ is the mass of that object, and
• $v$ is the speed of that object.

Important: Joule ($\rm J$) is the standard unit for energy. The formula for $\rm KE$ requires two inputs: mass and speed. The standard unit of mass is $\rm kg$ while the standard unit for speed is $\rm m \cdot s^{-1}$. If both inputs are in standard units, then the output (kinetic energy) will also be in the standard unit (that is: joules,

Convert the unit of the arrow's mass to standard unit:

$m = 63\; \rm g = 0.063\; \rm kg$.

Initial $\rm KE$ of this arrow:

$\begin{aligned}\mathrm{KE} &= \frac{1}{2} \, m \cdot v^2 \\ &= \frac{1}{2}\times 0.063\; \rm kg \times \left(\rm 28 \; m \cdot s^{-1}\right)^2 \\ &\approx 24.696\; \rm J\end{aligned}$.

That's the same as the energy output of this bow. Hence, the efficiency of energy transfer will be:

$\displaystyle \frac{24.696\; \rm J}{34.8\; \rm J} \times 100\% \approx 71\%$.