A body of 100 kg is placed on a truck. The coefficient of static friction between the body and the truck is 0.2. The truck suddenly decreases its speed from 90 km/hr to 36 km/hr in 5 second. Then

Options

A. The block does not move
B. the block slips forward and hits the driver's cabin
C. block shifts backward
D. nothing can be said about the block

Explanation

To solve the problem, we need to determine whether the body on the truck will slip or remain stationary as the truck decelerates. We will do this by calculating the deceleration of the truck and comparing it with the maximum static frictional force that can act on the body.

### Step-by-step Solution:

1. **Convert velocities from km/hr to m/s**:
- Initial velocity (U) = 90 km/hr = \( \frac{90 \times 1000}{3600} = 25 \, m/s \)
- Final velocity (V) = 36 km/hr = \( \frac{36 \times 1000}{3600} = 10 \, m/s \)

2. **Calculate the deceleration (a) of the truck**:
- We can use the formula for acceleration:
\[ a = \frac{V - U}{t} \]
- Substituting the values:
\[ a = \frac{10 \, m/s - 25 \, m/s}{5 \, s} = \frac{-15 \, m/s}{5 \, s} = -3 \, m/s^2 \]
- The negative sign indicates deceleration.

3. **Calculate the maximum static frictional force (F_friction)**:
- The formula for static friction is:
\[ F_{\text{friction}} = \mu_s \cdot N \]
- Where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force. For a horizontal surface, \( N = mg \):
\[ N = 100 \, kg \cdot 9.8 \, m/s^2 = 980 \, N \]
- Therefore, the maximum static frictional force is:
\[ F_{\text{friction}} = 0.2 \cdot 980 \, N = 196 \, N \]

4. **Calculate the force required to keep the body stationary (F_required)**:
- The force required to keep the body from slipping due to the truck's deceleration is given by:
\[ F_{\text{required}} = m \cdot a \]
- Substituting the values:
\[ F_{\text{required}} = 100 \, kg \cdot 3 \, m/s^2 = 300 \, N \]

5. **Compare the forces**:
- We have:
- Maximum static frictional force \( F_{\text{friction}} = 196 \, N \)
- Required force to keep the body stationary \( F_{\text{required}} = 300 \, N \)

6. **Conclusion**:
- Since \( F_{\text{required}} (300 \, N) > F_{\text{friction}} (196 \, N) \), the body will slip forward.