A body of mass 5kg is under the action of 50N on the horizontal surface. If coefficient of friction in between the surfaces is one, the distance it travels in 3 s is

Options

A. 2 m/s
B. 1.5 m
C. 0.9 m
D. 6 m

Answer

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

### Step 1: Identify the forces acting on the body
The body of mass \( m = 5 \, \text{kg} \) is subjected to a force \( F = 50 \, \text{N} \) in the horizontal direction. The frictional force must also be calculated since it opposes the motion.

### Step 2: Calculate the normal force
The normal force \( N \) acting on the body is equal to the weight of the body since it is on a horizontal surface.
\[ N = mg = 5 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 49 \, \text{N} \]

### Step 3: Calculate the frictional force
The frictional force \( f \) can be calculated using the formula:
\[ f = \mu N \]
where \( \mu = 1 \) (the coefficient of friction). Therefore,
\[ f = 1 \times 49 \, \text{N} = 49 \, \text{N} \]

### Step 4: Determine the net force acting on the body
The net force \( F_{\text{net}} \) acting on the body can be calculated by subtracting the frictional force from the applied force:
\[ F_{\text{net}} = F - f = 50 \, \text{N} - 49 \, \text{N} = 1 \, \text{N} \]

### Step 5: Calculate the acceleration of the body
Using Newton's second law, \( F = ma \), we can find the acceleration \( a \):
\[ F_{\text{net}} = ma \implies 1 \, \text{N} = 5 \, \text{kg} \cdot a \]
\[ a = \frac{1 \, \text{N}}{5 \, \text{kg}} = 0.2 \, \text{m/s}^2 \]

### Step 6: Calculate the distance traveled in 3 seconds
Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where \( u = 0 \) (initial velocity), \( a = 0.2 \, \text{m/s}^2 \), and \( t = 3 \, \text{s} \):
\[ s = 0 \times 3 + \frac{1}{2} \times 0.2 \times (3)^2 \]
\[ s = 0 + \frac{1}{2} \times 0.2 \times 9 = 0.1 \times 9 = 0.9 \, \text{m} \]

### Final Answer
The distance traveled by the body in 3 seconds is \( 0.9 \, \text{m} \).

---