If in pervious problem, an additional force `F' = 100N` is applied in vertical direction as shown in figure The friction force acting on the block is

Options

A. `zero`
B. `10 N`
C. `20 N`
D. `5 N`

Explanation

To solve the problem of finding the friction force acting on the block when an additional force \( F' = 100 \, \text{N} \) is applied in the vertical direction, we can follow these steps:

### Step 1: Identify the forces acting on the block
In this scenario, we have:
- The weight of the block \( mg \) acting downward.
- The additional force \( F' = 100 \, \text{N} \) acting downward.
- The normal force \( N \) acting upward.
- The applied horizontal force \( F = 5 \, \text{N} \).

### Step 2: Write down the equation of motion in the vertical direction
Since the lift is in free fall, the acceleration \( a \) of the block is equal to \( g \) (acceleration due to gravity). Therefore, we can write the equation of motion in the vertical direction as:
\[ F' + mg - N = ma \]
Substituting \( a = g \):
\[ 100 + mg - N = mg \]

### Step 3: Simplify the equation
Rearranging the equation gives:
\[ N = 100 \, \text{N} \]
This means that the normal force \( N \) acting on the block is equal to the additional force applied downward.

### Step 4: Calculate the friction force
The friction force \( f \) can be calculated using the formula:
\[ f = \mu N \]
Given that the coefficient of friction \( \mu = 0.1 \):
\[ f = 0.1 \times 100 \, \text{N} = 10 \, \text{N} \]

### Step 5: Determine the actual friction force
Since the applied horizontal force \( F = 5 \, \text{N} \) is less than the limiting friction force \( f = 10 \, \text{N} \), the friction force will equal the applied force:
\[ f_{\text{actual}} = F = 5 \, \text{N} \]

### Final Answer
The friction force acting on the block is \( 5 \, \text{N} \).

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