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 solve the problem, we need to find the friction force acting on a block of mass 15 kg placed on a trolley that accelerates with an acceleration of \(0.5 \, \text{m/s}^2\). The coefficient of friction between the block and the trolley is given as \(0.18\).

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block**:
- The block experiences two main forces: the gravitational force (weight) acting downwards and the normal force acting upwards. Additionally, as the trolley accelerates, a pseudo force acts on the block in the opposite direction of the trolley's acceleration.

2. **Calculate the Weight of the Block**:
\[ W = mg \]
where \(m = 15 \, \text{kg}\) and \(g = 9.8 \, \text{m/s}^2\) (approximated to \(10 \, \text{m/s}^2\) for simplicity).
\[ W = 15 \, \text{kg} \times 10 \, \text{m/s}^2 = 150 \, \text{N} \]

3. **Determine the Normal Force**:
- In this case, the normal force \(N\) is equal to the weight of the block since there are no vertical accelerations:
\[ N = W = 150 \, \text{N} \]

4. **Calculate the Maximum Frictional Force**:
\[ F_{\text{max}} = \mu_k \times N \]
where \(\mu_k = 0.18\).
\[ F_{\text{max}} = 0.18 \times 150 \, \text{N} = 27 \, \text{N} \]

5. **Calculate the Pseudo Force**:
- The pseudo force \(F_{\text{pseudo}}\) acting on the block due to the acceleration of the trolley is given by:
\[ F_{\text{pseudo}} = m \times a \]
where \(a = 0.5 \, \text{m/s}^2\).
\[ F_{\text{pseudo}} = 15 \, \text{kg} \times 0.5 \, \text{m/s}^2 = 7.5 \, \text{N} \]

6. **Determine the Friction Force**:
- The friction force \(F_f\) must balance the pseudo force to keep the block at rest relative to the trolley:
\[ F_f = F_{\text{pseudo}} = 7.5 \, \text{N} \]

### Final Answer:
The friction force acting on the block is \(7.5 \, \text{N}\).

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