Question

A particle moves under the effect of a force \( F = Cx \) from \( x = 0 \) to \( x = x_1 \). The work done in the process is

• A. \( C x_1^2 \)
• B. \( \frac{1}{2} C x_1^2 \)
• C. \( C x_1 \)
• D. Zero

Hint:

Graphically, work done is the area under the Force-Displacement (\( F \text{ vs } x \)) curve. For \( F = Cx \), the graph is a triangle with base \( x_1 \) and height \( Cx_1 \). Area = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} C x_1^2 \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Work done by a variable force is calculated by integrating the force over the displacement. Since the force \( F = Cx \) depends on position \( x \), it is a variable force.

Step 2: Key Formula or Approach:
The formula for work done \( W \) by a variable force is: \[ W = \int_{x_{initial}}^{x_{final}} F \, dx \]

Step 3: Detailed Explanation:
Substitute \( F = Cx \), and the limits from \( 0 \) to \( x_1 \): \[ W = \int_{0}^{x_1} Cx \, dx \] Factor out the constant \( C \): \[ W = C \int_{0}^{x_1} x \, dx \] Apply the power rule for integration \( \int x^n dx = \frac{x^{n+1}}{n+1} \): \[ W = C \left[ \frac{x^2}{2} \right]_{0}^{x_1} \] \[ W = C \left( \frac{x_1^2}{2} - \frac{0^2}{2} \right) \] \[ W = \frac{1}{2} C x_1^2 \]

Step 4: Final Answer
The work done in the process is \( \frac{1}{2} C x_1^2 \).