Chapter 6, Problem 38P

(a)

To determine

The kinetic energy of the bullet as it leaves the barrel.

Introduction: The kinetic energy of the bullet as it leaves the barrel is $4.56\text{kJ}$.

Explanation of Solution

Given Information:

The mass of a bullet is $15.0\text{g}$, the final speed of the bullet is $780\text{m}/\text{s}$ and the length of the barrel is $72.0\text{cm}$.

Formula to calculate the kinetic energy of the bullet is,

$K_{\text{f}}=\frac{1}{2}m{v}_{\text{f}}^{\text{2}}$

• $K_{\text{f}}$ is the kinetic energy of the bullet as it leaves the barrel.
• $m$ is the mass of the bullet.
• $v_{\text{f}}$ is the speed of the bullet as it leaves the barrel.

Substitute $15.0\text{g}$ for $m$ and $780\text{m}/\text{s}$ for $v_{\text{f}}$ to find $K_{\text{f}}$.

$\begin{array}{c}{K}_{\text{f}}=\frac{1}{2}\times \left(15.0g\times \frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\times {\left(780\text{m}/\text{s}\right)}^2\\ =4.56\text{kJ}\end{array}$

Conclusion:

Therefore, kinetic energy of the bullet as it leaves the barrel is $4.56\text{kJ}$.

(b)

To determine

The net work done on the bullet using work-kinetic energy theorem.

Answer to Problem 38P

The net work done on the bullet using work-kinetic energy theorem is $4.56\text{kJ}$.

Explanation of Solution

Given Information:

The mass of a bullet is $15.0\text{g}$, the final speed of the bullet is $780\text{m}/\text{s}$ and the length of the barrel is $72.0\text{cm}$.

From the work energy theorem, net work done is equal to the change in the kinetic energy of bullet.

$W=K_{\text{f}}-K_{\text{i}}$

• $W$ is the net work done on the bullet.
• $K_{\text{i}}$ is the initial kinetic energy of the bullet.

As the bullet starts from rest therefore, initial kinetic energy is zero.

Substitute $0\text{J}$ for $K_{\text{i}}$ and $4.56\text{kJ}$ for $K_{\text{f}}$ to find $W$.

$\begin{array}{c}W=4.56\text{kJ}-0\text{J}\\ =4.56\text{kJ}\end{array}$

Conclusion:

Therefore, the net work done on the bullet using work-kinetic energy theorem is $4.56\text{kJ}$.

(c)

To determine

The magnitude of the average net force that acted on the bullet in the barrel.

Answer to Problem 38P

The magnitude of the average net force that acted on the bullet in the barrel is $6.34\text{kN}$.

Explanation of Solution

Given Information:

The mass of a bullet is $15.0\text{g}$, the final speed of the bullet is $780\text{m}/\text{s}$ and the length of the barrel is $72.0\text{cm}$.

Formula to calculate the force acted on the bullet is,

$F=\frac{W}{l}$

• $F$ is the force acted on the bullet.
• $l$ is the length of the barrel.

Substitute $4.56\text{kJ}$ for $W$ and $72.0\text{cm}$ for $l$ to find $F$.

$\begin{array}{c}F=\frac{4.56\text{kJ}}{72.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}}\\ =6.34\text{kN}\end{array}$

Conclusion:

Therefore, the magnitude of the average net force that acted on the bullet in the barrel is $6.34\text{kN}$.

(d)

To determine

The constant acceleration of the bullet that start from a rest.

Answer to Problem 38P

The constant acceleration of the bullet that start from a rest is $422.5\text{km}/\text{s}$.

Explanation of Solution

Given Information:

The mass of a bullet is $15.0\text{g}$, the final speed of the bullet is $780\text{m}/\text{s}$ and the length of the barrel is $72.0\text{cm}$.

Formula to calculate the acceleration of the bullet using Newton’s law is,

$v_{\text{f}}^{\text{2}}=v_{\text{i}}^{\text{2}}+2al$

• $v_{\text{i}}$ is the initial velocity of the bullet.

Substitute $0\text{m}/\text{s}$ for $v_{\text{i}}$, $780\text{m}/\text{s}$ for $v_{\text{f}}$ and $72.0\text{cm}$ for $l$ to find $a$.

$\begin{array}{c}{\left(780\text{m}/\text{s}\right)}^2={\left(0\text{m}/\text{s}\right)}^2+2a\times \left(72.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ 608400{\text{m}}^2/{\text{s}}^2=\left(1.44\text{m}\right)a\\ a=422500\text{m}/\text{s}\times \frac{{10}^{-3}\text{km}/\text{s}}{1\text{m}/\text{s}}\\ =422.5\text{km}/\text{s}\end{array}$

Conclusion:

Therefore, the constant acceleration of the bullet that start from a rest is $422.5\text{km}/\text{s}$.

(e)

To determine

The net force acted on the bullet during acceleration.

Answer to Problem 38P

The net force acted on the bullet during acceleration is $6.34\text{kN}$.

Explanation of Solution

Given Information:

The mass of a bullet is $15.0\text{g}$, the final speed of the bullet is $780\text{m}/\text{s}$ and the length of the barrel is $72.0\text{cm}$.

Formula to calculate the net force acted on the bullet is,

$F=ma$

Substitute $15.0\text{g}$ for $m$ and $422.5\text{km}/\text{s}$ for $a$ to find $F$.

$\begin{array}{c}F=15.0\text{g}\times \frac{{10}^{-2}\text{kg}}{1\text{g}}\times 422.5\text{km}/\text{s}\times \frac{{10}^3\text{m}/\text{s}}{1\text{km}/\text{s}}\\ =6.34\text{kN}\end{array}$

Conclusion:

Therefore, the net force acted on the bullet during acceleration is $6.34\text{kN}$.

(f)

To determine

The conclusion that coming from the results of part (c) and (e).

Introduction: The work energy theorem relates the work done by the body to its energy and the Newton’s law equation is valid for the constant acceleration.

Explanation of Solution

Given Information:

The mass of a bullet is $15.0\text{g}$, the final speed of the bullet is $780\text{m}/\text{s}$ and the length of the barrel is $72.0\text{cm}$.

In part (c), the magnitude of the average net force that acted on the bullet in the barrel comes out to be $6.34\text{kN}$. While the net force acted on the bullet during acceleration in part (e) is $6.34\text{kN}$. Since the both the values comes out to be same hence, both the theories relate to each other.

Conclusion:

Therefore, both the theories work energy theorem and Newton’ equation concept satisfied to each other