Question:medium

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

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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 = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} C x_1^2 \).
Updated On: Apr 16, 2026
  • \( C x_1^2 \)
  • \( \frac{1}{2} C x_1^2 \)
  • \( C x_1 \)
  • Zero
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The Correct Option is B

Solution and Explanation

To solve the given problem of calculating the work done by a force as a particle moves from \( x = 0 \) to \( x = x_1 \), we need to understand the concept of work done by a variable force.

The force acting on the particle is given by \( F = Cx \), where \( C \) is a constant and \( x \) is the position of the particle.

The work done by a variable force when moving an object from position \( x_0 \) to position \( x_1 \) is given by the integral:

\(W = \int_{x_0}^{x_1} F(x) \, dx\)

In this case, we have:

\(W = \int_{0}^{x_1} Cx \, dx\)

Carrying out this integration:

\(W = C \int_{0}^{x_1} x \, dx = C \left[ \frac{x^2}{2} \right]_{0}^{x_1}\)

Evaluating the integral, we get:

\(W = C \left( \frac{x_1^2}{2} - \frac{0^2}{2} \right)\)

\(W = C \cdot \frac{x_1^2}{2}\)

Therefore, the work done by the force as the particle moves from \( x = 0 \) to \( x = x_1 \) is:

\(\frac{1}{2} C x_1^2\)

This matches the given correct answer option: \( \frac{1}{2} C x_1^2 \).

Conclusion: The correct answer is \(\frac{1}{2} C x_1^2\). This is based on applying the concept of work done by a variable force and accurately evaluating the integral for this specific force function.

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