A body of mass 2 kg is being dragged with uniform velocity of 2 m/s on a rough horizontal plane. The coefficient of friction between the body and the surface is 0.20. The amount of heat generated in 5 sec is `(J = 4.2 "joule /cal" and g = 9.8m//s^2)`

Options

A. 9.33 cal
B. 10.21 cal
C. 12.67 cal
D. 13.34 cal

Explanation

To solve the problem step by step, we need to determine the amount of heat generated when a body is dragged on a rough surface. Here’s how we can do that:

### Step 1: Identify the given data
- Mass of the body (m) = 2 kg
- Coefficient of friction (μ) = 0.20
- Velocity (v) = 2 m/s
- Time (t) = 5 seconds
- Acceleration due to gravity (g) = 9.8 m/s²

### Step 2: Calculate the normal force (N)
Since the body is on a horizontal plane, the normal force (N) is equal to the weight of the body:
\[ N = m \cdot g \]
\[ N = 2 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 19.6 \, \text{N} \]

### Step 3: Calculate the frictional force (F_f)
The frictional force can be calculated using the formula:
\[ F_f = \mu \cdot N \]
\[ F_f = 0.20 \cdot 19.6 \, \text{N} = 3.92 \, \text{N} \]

### Step 4: Calculate the displacement (s) during the time interval
Displacement can be calculated using the formula:
\[ s = v \cdot t \]
\[ s = 2 \, \text{m/s} \cdot 5 \, \text{s} = 10 \, \text{m} \]

### Step 5: Calculate the work done by the frictional force (W)
The work done by the frictional force is given by:
\[ W = F_f \cdot s \]
\[ W = 3.92 \, \text{N} \cdot 10 \, \text{m} = 39.2 \, \text{J} \]

### Step 6: Convert work done to heat generated
Since the work done against friction is converted to heat, the heat generated (Q) is:
\[ Q = W = 39.2 \, \text{J} \]

### Step 7: Convert joules to calories
To convert joules to calories, we use the conversion factor \( 1 \, \text{cal} = 4.2 \, \text{J} \):
\[ Q_{\text{cal}} = \frac{Q}{4.2} \]
\[ Q_{\text{cal}} = \frac{39.2 \, \text{J}}{4.2 \, \text{J/cal}} \approx 9.33 \, \text{cal} \]

### Final Answer
The amount of heat generated in 5 seconds is approximately **9.33 calories**.

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