A block of mass 15 kg is placed on a long trolley . The coefficient of friction between the block and the trolley is `0.18`. The trolley accelerates from rest with `0.5ms^(-2)` for 20s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground (b) as observer moving with the trolley.

Explanation

To solve the problem, we will analyze the motion of the block on the trolley from two perspectives: (a) a stationary observer on the ground and (b) an observer moving with the trolley.

### Step 1: Analyze the motion of the trolley
The trolley accelerates from rest with an acceleration of \(0.5 \, \text{m/s}^2\) for \(20 \, \text{s}\).

- **Final velocity of the trolley (v)** can be calculated using the formula:
\[ v = u + at \]
where \(u = 0\) (initial velocity), \(a = 0.5 \, \text{m/s}^2\), and \(t = 20 \, \text{s}\).

\[ v = 0 + (0.5 \times 20) = 10 \, \text{m/s} \]

### Step 2: Calculate the distance traveled by the trolley during acceleration
The distance \(s\) traveled by the trolley during the acceleration can be calculated using the formula:
\[ s = ut + \frac{1}{2}at^2 \]
Substituting the values:
\[ s = 0 \times 20 + \frac{1}{2} \times 0.5 \times (20)^2 = \frac{1}{2} \times 0.5 \times 400 = 50 \, \text{m} \]

### Step 3: Analyze the forces acting on the block
The block experiences a frictional force due to its contact with the trolley. The maximum static frictional force \(f_s\) can be calculated as:
\[ f_s = \mu_s \cdot m \cdot g \]
where:
- \(\mu_s = 0.18\) (coefficient of friction),
- \(m = 15 \, \text{kg}\) (mass of the block),
- \(g = 10 \, \text{m/s}^2\) (acceleration due to gravity).

Calculating \(f_s\):
\[ f_s = 0.18 \cdot 15 \cdot 10 = 27 \, \text{N} \]

### Step 4: Determine if the block will slide
The force required to accelerate the block along with the trolley can be calculated as:
\[ F = m \cdot a = 15 \cdot 0.5 = 7.5 \, \text{N} \]
Since the maximum static frictional force \(f_s = 27 \, \text{N}\) is greater than the force required to accelerate the block \(F = 7.5 \, \text{N}\), the block will not slide off the trolley during the acceleration phase.

### Step 5: Motion of the block as viewed by a stationary observer on the ground
- **During acceleration (first 20 seconds)**: The block will accelerate along with the trolley at \(0.5 \, \text{m/s}^2\) and will cover a distance of \(50 \, \text{m}\).
- **After 20 seconds**: The trolley moves with a constant velocity of \(10 \, \text{m/s}\). The block will also move with this constant velocity since it does not slide.

### Step 6: Motion of the block as viewed by an observer moving with the trolley
- **During acceleration**: The observer on the trolley will see the block at rest relative to them since both the block and the trolley are accelerating together.
- **After 20 seconds**: The observer will still see the block at rest since both are moving at the same constant velocity.

### Summary of Motion
- **Stationary observer**: The block accelerates for 20 seconds and then moves with a uniform velocity.
- **Observer on the trolley**: The block remains at rest relative to the trolley throughout the motion.