A block of mass `5kg` is placed at rest on a table of rough surface.Now,if a force of `30N` is applied in the direction parallel to surface of the table,the block slides through a distance of `50m` in an interval of time `10s.` Coefficient of kinetic friction is (given, `g=10ms^(-2)` )

Options

A. `0.50`
B. `0.60`
C. `0.75`
D. `0.25`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the forces acting on the block
The forces acting on the block are:
1. The applied force \( F = 30 \, \text{N} \) (in the direction of motion).
2. The frictional force \( f_k = \mu_k N \) (opposing the motion).
3. The normal force \( N = mg \) (acting perpendicular to the surface).

Given:
- Mass of the block \( m = 5 \, \text{kg} \)
- Gravitational acceleration \( g = 10 \, \text{m/s}^2 \)

Calculating the normal force:
\[ N = mg = 5 \, \text{kg} \times 10 \, \text{m/s}^2 = 50 \, \text{N} \]

### Step 2: Write the equation for net force
The net force \( F_{\text{net}} \) acting on the block can be expressed as:
\[ F_{\text{net}} = F - f_k \]
Substituting the expression for frictional force:
\[ F_{\text{net}} = 30 \, \text{N} - \mu_k N \]
\[ F_{\text{net}} = 30 \, \text{N} - \mu_k \times 50 \, \text{N} \]

### Step 3: Use Newton's second law to relate net force to acceleration
According to Newton's second law:
\[ F_{\text{net}} = ma \]
Thus, we can write:
\[ 30 - 50\mu_k = 5a \]

### Step 4: Determine the acceleration using the distance and time
We know the block slides through a distance \( s = 50 \, \text{m} \) in a time \( t = 10 \, \text{s} \). We can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the block starts from rest, \( u = 0 \):
\[ 50 = 0 + \frac{1}{2} a (10^2) \]
\[ 50 = 50a \]
\[ a = 1 \, \text{m/s}^2 \]

### Step 5: Substitute the acceleration back into the net force equation
Now we substitute \( a = 1 \, \text{m/s}^2 \) back into the net force equation:
\[ 30 - 50\mu_k = 5 \times 1 \]
\[ 30 - 50\mu_k = 5 \]
\[ 30 - 5 = 50\mu_k \]
\[ 25 = 50\mu_k \]
\[ \mu_k = \frac{25}{50} = 0.5 \]

### Final Answer
The coefficient of kinetic friction \( \mu_k \) is \( 0.5 \).

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