A block of mass 15kg is placed on a long trolley. The cofficient of fricition between the block and trolley is 0.18. The trolley accelerates from rest with `0.5m//s^(2)` for 20s. then what is the friction force:-

Options

A. 3.5N
B. 133.3N
C. 7.5N
D. N.O.T.

Explanation

To find the friction force acting on the block placed on the trolley, we can follow these steps:

### Step 1: Identify the Given Data
- Mass of the block (m) = 15 kg
- Coefficient of friction (μ) = 0.18
- Acceleration of the trolley (a) = 0.5 m/s²

### Step 2: Calculate the Required Friction Force
The required friction force (F_friction) to keep the block moving with the trolley can be calculated using Newton's second law. The friction force must provide the necessary force to accelerate the block along with the trolley.

Using the formula:
\[ F_{\text{friction}} = m \cdot a \]

Substituting the values:
\[ F_{\text{friction}} = 15 \, \text{kg} \cdot 0.5 \, \text{m/s}^2 \]
\[ F_{\text{friction}} = 7.5 \, \text{N} \]

### Step 3: Calculate the Normal Force
The normal force (N) acting on the block is equal to its weight since there are no vertical accelerations. This can be calculated as:
\[ N = m \cdot g \]
where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \).

Substituting the values:
\[ N = 15 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \]
\[ N = 147.15 \, \text{N} \]

### Step 4: Calculate the Limiting Friction
The limiting friction (F_limiting) can be calculated using the coefficient of friction and the normal force:
\[ F_{\text{limiting}} = \mu \cdot N \]

Substituting the values:
\[ F_{\text{limiting}} = 0.18 \cdot 147.15 \, \text{N} \]
\[ F_{\text{limiting}} = 26.46 \, \text{N} \]

### Step 5: Compare Required Friction Force and Limiting Friction
Now, we compare the required friction force with the limiting friction:
- Required friction force = 7.5 N
- Limiting friction = 26.46 N

Since the required friction force (7.5 N) is less than the limiting friction (26.46 N), the block will not slip on the trolley.

### Conclusion
The friction force acting on the block is 7.5 N.

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