An iron block of mass 5 kg is kept on a trolley. If the trolley is being pushed with an acceleration of 5 m/`s^(2)`, What will be the force of friction between the block and the trolley surface (Take the coefficient of static friction between the block and the surface to be 0.8)

Options

A. Zero
B. 5 N
C. 4 N
D. 25 N

Explanation

To solve the problem, we need to determine the force of friction between the iron block and the trolley surface when the trolley is being pushed with a certain acceleration. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Mass of the block (m) = 5 kg
- Acceleration of the trolley (a) = 5 m/s²
- Coefficient of static friction (μ) = 0.8
- Acceleration due to gravity (g) = 10 m/s² (approximated)

### Step 2: Calculate the normal force (N)
The normal force acting on the block is equal to the weight of the block, which can be calculated using the formula:
\[ N = m \cdot g \]
Substituting the values:
\[ N = 5 \, \text{kg} \cdot 10 \, \text{m/s}² = 50 \, \text{N} \]

### Step 3: Calculate the maximum static frictional force (f_max)
The maximum static frictional force can be calculated using the formula:
\[ f_{\text{max}} = \mu \cdot N \]
Substituting the known values:
\[ f_{\text{max}} = 0.8 \cdot 50 \, \text{N} = 40 \, \text{N} \]

### Step 4: Calculate the force due to the trolley motion on the block (F)
The force exerted by the trolley on the block can be calculated using Newton's second law:
\[ F = m \cdot a \]
Substituting the values:
\[ F = 5 \, \text{kg} \cdot 5 \, \text{m/s}² = 25 \, \text{N} \]

### Step 5: Compare the force due to trolley motion with the maximum static frictional force
Since the force due to the trolley motion (25 N) is less than the maximum static frictional force (40 N), the block will not slide on the trolley. This means that the frictional force will equal the force due to the trolley motion.

### Step 6: Conclusion
The force of friction between the block and the trolley surface is:
\[ f = F = 25 \, \text{N} \]

Thus, the force of friction between the block and the trolley surface is **25 N**.

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