A body of mass 5 kg rests on a rough horizontal surface of coefficient of friction 0.2. The body is pulled through a distance of 10 m by a horizontal force of 25 N . The kinetic energy acquired by it is `(g=10ms^(2))`

Options

A. 330 J
B. 150 J
C. 100 J
D. 50 J

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the Normal Force (N)
The normal force acting on the body can be calculated using the formula:
\[ N = mg \]
Where:
- \( m = 5 \, \text{kg} \) (mass of the body)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

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

### Step 2: Calculate the Force of Kinetic Friction (f_k)
The force of kinetic friction can be calculated using the formula:
\[ f_k = \mu N \]
Where:
- \( \mu = 0.2 \) (coefficient of friction)

Calculating:
\[ f_k = 0.2 \times 50 \, \text{N} = 10 \, \text{N} \]

### Step 3: Calculate the Work Done by the Applied Force (W_applied)
The work done by the applied force can be calculated using the formula:
\[ W_{\text{applied}} = F \times d \]
Where:
- \( F = 25 \, \text{N} \) (applied force)
- \( d = 10 \, \text{m} \) (distance moved)

Calculating:
\[ W_{\text{applied}} = 25 \, \text{N} \times 10 \, \text{m} = 250 \, \text{J} \]

### Step 4: Calculate the Work Done by Friction (W_friction)
The work done by friction is given by:
\[ W_{\text{friction}} = -f_k \times d \]
The negative sign indicates that friction does work against the direction of motion.

Calculating:
\[ W_{\text{friction}} = -10 \, \text{N} \times 10 \, \text{m} = -100 \, \text{J} \]

### Step 5: Calculate the Net Work Done (W_net)
The net work done on the body is the sum of the work done by the applied force and the work done by friction:
\[ W_{\text{net}} = W_{\text{applied}} + W_{\text{friction}} \]

Calculating:
\[ W_{\text{net}} = 250 \, \text{J} - 100 \, \text{J} = 150 \, \text{J} \]

### Step 6: Calculate the Kinetic Energy Acquired (K)
According to the work-energy theorem:
\[ W_{\text{net}} = \Delta K \]
Since the body starts from rest, the change in kinetic energy is equal to the final kinetic energy:
\[ K = W_{\text{net}} = 150 \, \text{J} \]

### Final Answer
The kinetic energy acquired by the body is:
\[ \boxed{150 \, \text{J}} \]

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