Chapter 8, Problem 78AP

(a)

To determine

The maximum speed of the needle.

Answer to Problem 78AP

The maximum speed of the needle is $\underset{\_}{21\text{m/s}}$.

Explanation of Solution

Consider the needle-spring-body as a system. Since there are resisting forces when the needle moves through the skin. Hence, the maximum speed of the needle when projected horizontally using a spring will be at the point on the patient body.

Consider, the spring-needle as an isolated system.

Since the total energy of the system is conserved from the point the spring is maximumly compressed to the point the needle is on the point on the body of the patient. Hence, the total kinetic and potential energy will be conserved.

Write the equation for conservation of energy

  $\Delta U_s+\Delta K=0$                                                                                                (I)

Here, $\Delta U_s$ is the change in spring potential energy and $\Delta K$ is the change in kinetic energy.

Write the expression for the change in spring potential energy

  $\Delta U_s=U_{sf}-U_{si}$                                                                                            (II)

Here, $U_s{}_f$ is the final potential energy of the spring and $U_{si}$ is the initial potential energy.

Write the equation for spring potential energy

  $U_s=\frac{1}{2}k{x}^2$

Here, $U_s$ is spring potential energy, $k$ is the spring constant and $x$ is the compressed length of the spring.

Since the spring is compressed initially and when the needle touches the skin spring has no potential energy.

Substitute $\frac{1}{2}k{x}^2$ for $U_{si}$ and $0$ for $U_{sf}$ in equation (II).

  $\Delta U_s=0-\left(\frac{1}{2}k{x}^2\right)$

Simplify the above equation.

  $\Delta U_s=-\frac{1}{2}k{x}^2$

Write the expression for the change in kinetic energy of the spring-needle system

  $\Delta K=K_f-K_i$                                                                                             (III)

Here, $\Delta K$ is the change in kinetic energy of the spring needle system, $K_f$ is the final kinetic energy and $K_i$ is the initial kinetic energy.

Since the needle is the one having the kinetic energy change when the needle moves from spring to patient body.

Write the equation for the kinetic energy of the needle

  $K=\frac{1}{2}m{v}^2$

Here, $K$ is the kinetic energy of the system, $m$ is the mass of the system and $v$ is the speed of the system.

Substitute  $\frac{1}{2}m{v}^2$ for $K_f$ and $0$ for $K_i$ in equation (III).

  $\Delta K=\left(\frac{1}{2}m{v}^2\right)-0$

Simplify the above equation.

  $\Delta K=\frac{1}{2}m{v}^2$

Substitute $-\frac{1}{2}k{x}^2$ for $\Delta U_s$ and $\frac{1}{2}m{v}^2$ for $\Delta K$ in equation (I).

  $\left(-\frac{1}{2}k{x}^2\right)+\left(\frac{1}{2}m{v}^2\right)=0$

Rearrange the above equation.

  $v=\sqrt{\frac{2\left(\frac{1}{2}k{x}^2\right)}{m}}$

Simplify the above equation.

  $v=x\sqrt{\frac{k}{m}}$                                                                                                     (IV)

Conclusion:

Substitute $375\text{N/m}$ for $k$, $8.10\text{cm}$ for $x$ and $5.60\text{g}$ for $m$ in equation (IV).

  $\begin{array}{c}v=\sqrt{\frac{\left(375\text{N/m}\right)}{\left(5.60\text{g}\right)}}\left(8.10\text{cm}\right)\\ =\left(8.10\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\sqrt{\frac{\left(375\text{N/m}\right)}{\left(5.60\text{g}\right)}\left(\frac{1000\text{g}}{1\text{kg}}\right)}\\ =\left(0.0810\text{m}\right)\sqrt{66964.28571\text{N}/\text{m}\cdot \text{kg}}\\ =21\text{m/s}\end{array}$

Thus, the maximum speed of the needle is $\underset{\_}{21\text{m/s}}$.

(b)

To determine

The speed at which the flange on the back end of the needle runs into a stop that is set to limit the penetration.

Answer to Problem 78AP

The speed at which the flange on the back end of the needle runs into a stop that is set to limit the penetration is $\underset{\_}{16.1\text{m/s}}$.

Explanation of Solution

Consider the needle-spring-body system as a system.

Since there are resisting forces when the needle moves through the skin.

Hence, the total energy of the needle while penetrating the skin is transformed into work done by the resisting forces.

Consider the needle-body as an isolated system.

Since the work done on the system is by internal resisting forces which are non-conservative.

Write the equation for conservation for energy

  $\Delta U+\Delta K=W$                                                                                              (V)

Here, $W$ is total work done by resisting forces.

Since the needle moves horizontally, hence there is no change in potential energy.

Write the expression for work done

  $W=W_1+W_2$

Here, $W_1$ is the work done by resisting force exerted by the skin, soft tissues and $W_2$ is the work done by resisting force exerted by organ.

Write the equation for work done by resisting force

  $W=-Fd$

Here, $F$ is the force in the opposite direction of displacement and $d$ is the magnitude of displacement.

Write the expression for work done using the above equation

  $W=-\left(F_1{d}_1+F_2{d}_2\right)$                                                                                     (VI)

Here, $F_1$ is the resisting force exerted by soft tissues, $d_1$ is the distance needle traveled under the force exerted by soft tissues, $F_2$ is the resisting force exerted by organ and $d_2$ is the distance needle traveled under the force exerted by organ.

Since the needle is the one having the kinetic energy change when the needle moves from spring to patient body.

Substitute $\frac{1}{2}m{v}^2$ for $K_f$ and $0$ for $K_i$ in equation (III).

  $\Delta K=\left(\frac{1}{2}m{v}^2\right)-0$

Simplify the above equation.

  $\Delta K=\frac{1}{2}m{v}^2$

Since the spring is compressed initially and when the needle touches the skin spring has no potential energy.

Substitute $\frac{1}{2}k{x}^2$ for $U_{si}$ and $0$ for $U_{sf}$ in equation (II).

  $\Delta U_s=0-\left(\frac{1}{2}k{x}^2\right)$

Simplify the above equation.

  $\Delta U_s=-\frac{1}{2}k{x}^2$

Substitute $-\frac{1}{2}k{x}^2$ for $\Delta U_s$ and $\frac{1}{2}m{v}^2$ for $\Delta K$ in equation (V).

  $\left(-\frac{1}{2}k{x}^2\right)+\left(\frac{1}{2}m{v}^2\right)=W$

Rearrange the above equation.

  $v=\sqrt{\frac{2\left(\frac{1}{2}k{x}^2\right)+2W}{m}}$

Simplify the above equation.

  $v=\sqrt{\frac{k{x}^2+2W}{m}}$                                                                                          (VII)

Conclusion:

Substitute $2.40\text{cm}$ for $d_1$, $7.60\text{N}$ for $F_1$, $3.50\text{cm}$ for $d_2$ and $9.20\text{N}$ for $F_2$ in equation (VI).

  $\begin{array}{l}W=-\left(\left(7.60\text{N}\right)\left(2.40\text{cm}\right)+\left(9.20\text{N}\right)\left(3.50\text{cm}\right)\right)\\ =-\left(\left(18.24\text{N}\cdot \text{cm}\right)+\left(32.2\text{N}\cdot \text{cm}\right)\right)\\ =-\left(50.44\text{N}\cdot \text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =-0.544\text{N}\cdot \text{m}\end{array}$

Substitute $375\text{N/m}$ for $k$, $8.10\text{cm}$ for $x$ and $5.60\text{g}$ for $m$, $-0.544\text{N}\text{m}$ for $W$ in equation (VII).

  $\begin{array}{c}v=\sqrt{\frac{\left(375\text{N/m}\right){\left(8.10\text{cm}\right)}^2+2\times \left(-0.504\text{N}\cdot \text{m}\right)}{\left(5.60\text{g}\right)}}\\ =\sqrt{\frac{\left(\text{24603}\text{.75 N}\cdot {\text{cm}}^{\text{2}}\right)\left(\frac{1{\text{m}}^2}{10000{\text{cm}}^2}\right)-\left(2\times 0.504\text{N}\cdot \text{m}\right)}{\left(5.60\text{g}\right)\left(\frac{1\text{kg}}{1000\text{g}}\right)}}\\ =\sqrt{259.35{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ =16.1\text{m/s}\end{array}$

Thus, the speed at which the flange on the back end of the needle runs into a stop that is set to limit the penetration is $\underset{\_}{16.1\text{m/s}}$.