A 10.0-g bullet traveling horizontally at 755 m/s strikes a stationary target and stops after penetrating 14.5 cm into the target. What is the average force of the target on the bullet ?

Options

A. `1.97xx10^(4)`N
B. `6.26xx10^(3)`N
C. `2.07xx10^(5)`N
D. `3.13xx10^(4)N`

Answer

To solve the problem, we will use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution

1. **Identify the Given Data:**
- Mass of the bullet, \( m = 10.0 \, \text{g} = 0.010 \, \text{kg} \) (convert grams to kilograms)
- Initial velocity of the bullet, \( v_i = 755 \, \text{m/s} \)
- Final velocity of the bullet, \( v_f = 0 \, \text{m/s} \) (since it comes to a stop)
- Penetration distance, \( d = 14.5 \, \text{cm} = 0.145 \, \text{m} \) (convert centimeters to meters)

2. **Calculate the Initial Kinetic Energy (KE_initial):**
\[ KE_{\text{initial}} = \frac{1}{2} m v_i^2 \]
Substituting the values:
\[ KE_{\text{initial}} = \frac{1}{2} \times 0.010 \, \text{kg} \times (755 \, \text{m/s})^2 \]
\[ KE_{\text{initial}} = 0.005 \times 570025 = 2850.125 \, \text{J} \]

3. **Calculate the Final Kinetic Energy (KE_final):**
\[ KE_{\text{final}} = 0 \, \text{J} \quad (\text{since the bullet stops}) \]

4. **Calculate the Work Done (W):**
The work done by the target on the bullet is equal to the change in kinetic energy:
\[ W = KE_{\text{final}} - KE_{\text{initial}} = 0 - 2850.125 = -2850.125 \, \text{J} \]
(The negative sign indicates that the work is done against the bullet's motion.)

5. **Use the Work-Energy Principle to Find the Average Force (F):**
The work done is also equal to the force multiplied by the distance over which it acts:
\[ W = F \cdot d \]
Rearranging for force gives:
\[ F = \frac{W}{d} \]
Substituting the values:
\[ F = \frac{-2850.125 \, \text{J}}{0.145 \, \text{m}} = -19656.034 \, \text{N} \]
(The negative sign indicates that the force is in the opposite direction to the bullet's motion.)

6. **Final Result:**
The average force of the target on the bullet is approximately:
\[ F \approx 19656.034 \, \text{N} \quad \text{or} \quad 1.9656 \times 10^4 \, \text{N} \]