Question

A force \( F = (2 + x) \) acts on a particle in x-direction, where \(F\) is in newton and \(x\) in metre. The work done during displacement from \(x=1\) m to \(x=2\) m is:

• A. \(2 \, J\)
• B. \(3.5 \, J\)
• C. \(4.5 \, J\)
• D. None of these

Hint:

Split integrals into simple parts for faster calculation!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Work done by a force is the integral of the force over the path of displacement.
Since the force in this problem is a function of position \(x\), we must perform integration to find the total work done.
Step 2: Key Formula or Approach:
Work done \(W\) is given by:
\[ W = \int_{x_{i}}^{x_{f}} F dx \]
Step 3: Detailed Explanation:
Given \(F = 2 + x\).
The limits of integration are from \(x_{i} = 1.0\text{ m}\) to \(x_{f} = 2.0\text{ m}\).
\[ W = \int_{1}^{2} (2 + x) dx \]
Perform the integration:
\[ W = \left[ 2x + \frac{x^{2}}{2} \right]_{1}^{2} \]
Evaluate at the limits:
\[ W = \left( 2(2) + \frac{2^{2}}{2} \right) - \left( 2(1) + \frac{1^{2}}{2} \right) \]
\[ W = (4 + 2) - (2 + 0.5) \]
\[ W = 6 - 2.5 = 3.5\text{ J} \]
Step 4: Final Answer:
The work done by the force is \(3.5\text{ J}\).