Question

The relationship between the applied force \( F(X) \) (in Newton) on a body and its displacement \( X \) (in metre) is given below. The total amount of work done in moving the body from \( X = 0 \) to \( X = 4 \, \text{m} \) is ............ Joule. 

Hint:

The work done by a variable force is the area under the force vs displacement graph. For linear segments, use the area of a rectangle or triangle as appropriate.

Answer 1

Correct Answer: 10.99 - 11.01

Step 1: Understanding the concept of work done. 
The work done \( W \) by a variable force is given by the area under the force-displacement graph. The graph provided has three parts: a constant force from \( 0 \) to \( 2 \, \text{m} \), a linearly increasing force from \( 2 \) to \( 4 \, \text{m} \), and a constant force at \( X = 4 \).

Step 2: Calculating the area under the graph. 
- From \( X = 0 \) to \( X = 2 \, \text{m} \), the force is constant at 2 N. The work done in this section is the area of a rectangle: \[ W_1 = \text{Force} \times \text{Displacement} = 2 \, \text{N} \times 2 \, \text{m} = 4 \, \text{J} \] - From \( X = 2 \) to \( X = 4 \, \text{m} \), the force increases linearly from 2 N to 4 N. The work done in this section is the area of a triangle: \[ W_2 = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 2 \, \text{m} \times (4 \, \text{N} - 2 \, \text{N}) = 2 \, \text{J} \]

Step 3: Total work done. 
The total work done is the sum of the work in both sections: \[ W = W_1 + W_2 = 4 \, \text{J} + 2 \, \text{J} = 10 \, \text{J} \]

Step 4: Conclusion. 
Thus, the total work done in moving the body from \( X = 0 \) to \( X = 4 \, \text{m} \) is \( 10 \, \text{J} \), and the correct answer is 10 Joules.