The average force necessary to stop a hammer with 25 N-sec momentum is 0.05 sec expressed in N is

Options

A. 500
B. 125
C. 50
D. 25

Explanation

To solve the problem of finding the average force necessary to stop a hammer with a momentum of 25 N·s in 0.05 seconds, we can use the formula derived from Newton's second law of motion:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Initial momentum (P_initial) = 25 N·s (Newton-seconds)
- Final momentum (P_final) = 0 N·s (since the hammer is stopped)
- Time interval (Δt) = 0.05 seconds

2. **Calculate the Change in Momentum (ΔP)**:
- Change in momentum (ΔP) = P_final - P_initial
- ΔP = 0 N·s - 25 N·s = -25 N·s

3. **Use the Formula for Average Force (F)**:
- According to Newton's second law, the average force can be calculated using the formula:
\[ F = \frac{\Delta P}{\Delta t} \]

4. **Substitute the Values into the Formula**:
- F = ΔP / Δt
- F = -25 N·s / 0.05 s

5. **Calculate the Average Force**:
- F = -25 / 0.05
- F = -500 N

6. **Interpret the Result**:
- The negative sign indicates that the force is applied in the opposite direction of the momentum (which is expected when stopping an object). Therefore, the average force necessary to stop the hammer is 500 N.

### Final Answer:
The average force necessary to stop the hammer is **500 N**.

---