A body of mass 5 kg is acted on by a net force F which varies with time t as shown in graph Fig. Then the net momentum is SI units gained by the body at the end of 10 seconds is

Options

A. 0
B. 100
C. 140
D. 200

Explanation

To find the net momentum gained by a body of mass 5 kg acted upon by a varying net force over a period of 10 seconds, we will follow these steps:

### Step 1: Understand the Relationship Between Force and Momentum
The relationship between force (F) and momentum (p) is given by the equation:
\[ F = \frac{dp}{dt} \]
This implies that the change in momentum is equal to the force applied over time:
\[ dp = F \, dt \]

### Step 2: Calculate the Total Change in Momentum
To find the total change in momentum over a time interval, we can integrate the force with respect to time:
\[ p = \int F \, dt \]
In the context of a force-time graph, the area under the graph represents the change in momentum.

### Step 3: Analyze the Force-Time Graph
From the problem, we have a force-time graph with three distinct areas:
1. A triangle from \( t = 0 \) to \( t = 4 \) seconds.
2. A rectangle from \( t = 4 \) to \( t = 8 \) seconds.
3. A triangle from \( t = 8 \) to \( t = 10 \) seconds.

### Step 4: Calculate the Area of Each Section
1. **Area of Triangle 1 (A1)**:
- Base = 4 seconds (from 0 to 4)
- Height = 20 N
- Area (A1) = \( \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 4 \times 20 = 40 \, \text{N·s} \)

2. **Area of Rectangle (A2)**:
- Width = 4 seconds (from 4 to 8)
- Height = 20 N
- Area (A2) = \( \text{Width} \times \text{Height} = 4 \times 20 = 80 \, \text{N·s} \)

3. **Area of Triangle 2 (A3)**:
- Base = 2 seconds (from 8 to 10)
- Height = 20 N
- Area (A3) = \( \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 2 \times 20 = 20 \, \text{N·s} \)

### Step 5: Calculate the Total Momentum
Now, we sum all the areas to find the total momentum:
\[ p = A1 + A2 + A3 = 40 + 80 + 20 = 140 \, \text{N·s} \]

### Step 6: Conclusion
The net momentum gained by the body at the end of 10 seconds is:
\[ \text{Net Momentum} = 140 \, \text{N·s} \]