A body of mass 10kg is under the action of 138N on the horizontal surface. If coefficient of friction in between the surfaces is one, the distance it travels in 10s in metre is

Options

A. 100
B. 200
C. 300
D. 500

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify the Forces Acting on the Body
The body has a mass \( m = 10 \, \text{kg} \) and is subjected to a force \( F = 138 \, \text{N} \) horizontally. The weight of the body can be calculated as:
\[ W = mg = 10 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 98 \, \text{N} \]

### Step 2: Determine the Normal Force
Since the body is on a horizontal surface and there are no vertical forces acting other than weight and the normal force, the normal force \( N \) is equal to the weight of the body:
\[ N = W = 98 \, \text{N} \]

### Step 3: Calculate the Frictional Force
The frictional force \( f \) can be calculated using the coefficient of friction \( \mu = 1 \):
\[ f = \mu N = 1 \times 98 \, \text{N} = 98 \, \text{N} \]

### Step 4: Determine the Net Force Acting on the Body
The net force \( F_{\text{net}} \) acting on the body can be found by subtracting the frictional force from the applied force:
\[ F_{\text{net}} = F - f = 138 \, \text{N} - 98 \, \text{N} = 40 \, \text{N} \]

### Step 5: Calculate the Acceleration of the Body
Using Newton's second law \( F = ma \), we can find the acceleration \( a \):
\[ F_{\text{net}} = ma \implies 40 \, \text{N} = 10 \, \text{kg} \times a \implies a = \frac{40 \, \text{N}}{10 \, \text{kg}} = 4 \, \text{m/s}^2 \]

### Step 6: Calculate the Distance Traveled in 10 Seconds
Using the second equation of motion, where the initial velocity \( u = 0 \):
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ s = 0 + \frac{1}{2} \times 4 \, \text{m/s}^2 \times (10 \, \text{s})^2 = \frac{1}{2} \times 4 \times 100 = 200 \, \text{m} \]

### Final Answer
The distance traveled by the body in 10 seconds is:
\[ \boxed{200 \, \text{m}} \]