The block moves 24 meters in the next 5 seconds after the man stops pushing.
The motion of the block can be analyzed using Newton's second law and the equations of motion.
1. **Determine Acceleration (a):**
Using Newton's second law \( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration. In this case, the applied force is 100 N.
\[ a = \frac{F}{m} \]
2. **Determine Initial Velocity (u):**
The block starts from rest, so the initial velocity \( u \) is 0 m/s.
3. **Use Equations of Motion:**
Using the equation of motion \( s = ut + \frac{1}{2}at^2 \), where \( s \) is the displacement, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.
- **First 5 Seconds:**
\[ s_1 = ut_1 + \frac{1}{2}a(t_1)^2 \]
Substituting the given values:
\[ s_1 = 0 \times 5 + \frac{1}{2}a(5)^2 \]
Solving for \( a \):
\[ 12 = \frac{1}{2}a(5)^2 \]
\[ a = \frac{12 \times 2}{5^2} \]
\[ a = \frac{24}{25} \, \text{m/s}^2 \]
Now that we have \( a \), we can find \( s_1 \) during the first 5 seconds.
- **Next 5 Seconds:**
The man stops pushing, so there is no force applied. The block moves with its acquired velocity from the first 5 seconds.
\[ v = u + at \]
\[ v = 0 + \frac{24}{25} \times 5 \]
Now, use \( s = ut + \frac{1}{2}at^2 \) again to find the displacement during the next 5 seconds.
\[ s_2 = ut_2 + \frac{1}{2}a(t_2)^2 \]
Substituting the values:
\[ s_2 = \frac{24}{25} \times 5 \times 5 + \frac{1}{2} \times 0 \times (5)^2 \]
Simplifying:
\[ s_2 = \frac{600}{25} \]
\[ s_2 = 24 \, \text{m} \]
**Result:**
The block moves 24 meters in the next 5 seconds after the man stops pushing.