Chapter 7, Problem 83P

(a)

To determine

Maximum speed of the needle.

Answer to Problem 83P

Maximum speed is $\underset{\_}{21\text{m}/\text{s}}$.

Explanation of Solution

Consider the needle is fired horizontally from a spring. Maximum speed occurs just after the needle leaves the spring, before entering the body.

Write the law of conservation of energy for this case.

  $K_i+U_i-E=K_f+U_f$        (I)

Here $K_i$ is the initial kinetic energy, $U_i$ is the initial elastic potential energy, $K_f$ is the final kinetic energy $U_f$ is the final elastic potential energy and $E$ is the work done by the spring.

Write the equation for initial and final kinetic energy.

  $\begin{array}{l}{K}_i=\frac{1}{2}m{v}_i{}^2\\ K_f=\frac{1}{2}m{v}_f{}^2\end{array}$        (II)

Here $m$ is the mass of the needle, $v_i$ is the initial velocity and $v_f$ is the final velocity.

Write down the equation for initial and elastic potential energy.

  $\begin{array}{l}{U}_i=\frac{1}{2}k{x}_i{}^2\\ U_f=\frac{1}{2}k{x}_f{}^2\end{array}$        (III)

Here $k$ is the spring constant, $x_i$ is the initial extension of the spring and $x_f$ is the final extension of the spring.

Write the equation for work done by the spring

  $E=f_{k}d$        (IV)

Here $E$ is the work done by the spring, $f_k$ is the force constant of the spring and $d$ is the distance moved into the tissue.

As the initial velocity s zero, $K_i=0$

As there is no penetration into the body, distance covered is zero. Then $E=0$

As there is no extension, $U_f=0$

Substitute these results in (I) and write for $v_f$.

  $\begin{array}{c}0+\frac{1}{2}k{x}_i{}^2-0=\frac{1}{2}m{v}_f{}^2+0\\ v_f{}^2=\frac{k{x}_i{}^2}{m}\\ v_f=\sqrt{\frac{k}{m}}{x}_i\end{array}$        (V)

Conclusion:

Substitute $375\text{N}/\text{m}$ for $k$, $5.6\text{g}$ for $m$ and $8.1\text{cm}$ for $x_i$ in (V)

  $\begin{array}{c}{v}_f=\sqrt{\frac{375\text{N}/\text{m}}{5.6\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)}}\left(8.1\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\\ =\sqrt{\frac{375\text{N}/\text{m}}{5.6\times {10}^{-3}\text{kg}}}\left(8.1\times {10}^{-2}\text{m}\right)\\ =20.96\text{m}/\text{s}\\ \approx 21\text{m}/\text{s}\end{array}$

Maximum speed is $\underset{\_}{21\text{m}/\text{s}}$.

(b)

To determine

Speed to limit the penetration to $5.9\text{cm}$.

Answer to Problem 83P

Speed should be $\underset{\_}{16\text{m}/\text{s}}$.

Explanation of Solution

The initial elastic potential energy is converted partially into internal energy in the organ and partially kinetic energy of the needle.

Write the energy conservation equation for this case.

  $K_i+U_i-E_i-E_f=K_f+U_f$        (VI)

Here $K_i$ is the initial kinetic energy, $U_i$ is the initial elastic potential energy, $K_f$ is the final kinetic energy $U_f$ is the final elastic potential energy $E_i$ is the initial work done by the spring and $E_f$ is the final work done by the spring.

Write the equation for initial and final work done by the spring

  $\begin{array}{l}{E}_i=f_{ki}{d}_i\\ E_f=f_{kf}{d}_f\end{array}$        (VII)

Here, $f_k{}_i$ is the initial force constant of the spring, $f_{kf}$ is the final force constant of the spring, $d_i$ is the initial distance moved into the tissue and $d_f$ is the final distance moved into the tissue.

As the initial velocity s zero

  $K_i=0$        (VIII)

As there is no extension,

  $U_f=0$        (IX)

Substitute (II), (III), (VII), (VIII) and (IX) in (VI) and write for $v_f$.

  $\begin{array}{c}\frac{1}{2}m{v}_f{}^2+0=0+\frac{1}{2}k{x}_i{}^2-f_{ki}{d}_i-f_{kf}{d}_f\\ v_f{}^2=\frac{2\left(\frac{1}{2}k{x}_i{}^2-f_{ki}{d}_i-f_{kf}{d}_f\right)}{m}\\ v_f=\sqrt{\frac{2\left(\frac{1}{2}k{x}_i{}^2-f_k{d}_i-f_k{d}_f\right)}{m}}\end{array}$        (X)

Conclusion:

Substitute $375\text{N}/\text{m}$ for $k$, $5.6\text{g}$ for $m$, $8.1\text{cm}$ for $x_i$, $7.6\text{N}$ for $f_{ki}$, $2.4\text{cm}$ for $d_i$, $9.2\text{N}$ for $f_{kf}$ and $3.5\text{cm}$ for $x_f$ in (X)

  $\begin{array}{c}{v}_f=\sqrt{\frac{2\left(\frac{1}{2}375\text{N}/\text{m}{\left(8.1\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2-\left(\left(7.6\text{N}\right)\left(2.4\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\right)-\left(\left(9.2\text{N}\right)\left(3.5\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\right)\right)}{5.6\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)}}\\ =\sqrt{\frac{2\left(\frac{1}{2}375\text{N}/\text{m}{\left(8.1\times {10}^{-2}\text{m}\right)}^2-\left(\left(7.6\text{N}\right)\left(2.4\times {10}^{-2}\text{m}\right)\right)-\left(\left(9.2\text{N}\right)\left(3.5\times {10}^{-2}\text{m}\right)\right)\right)}{5.6\times {10}^{-3}\text{kg}}}\\ =16\text{m}/\text{s}\end{array}$

Speed should be $\underset{\_}{16\text{m}/\text{s}}$.