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Question: A particle moves along the x-axis from x = \(x_1\) to x = \(x_2\) under the influence of a force giv...

A particle moves along the x-axis from x = x1x_1 to x = x2x_2 under the influence of a force given by F = 2x. Then work done in the process is
A. Zero
B. x22−x12x_2^2 - x_1^2
C. 2x2(x2−x1)2{x_2}\left( {{x_2} - {x_1}} \right)
D. 2x1(x1−x2)2{x_1}\left( {{x_1} - {x_2}} \right)

Explanation

Solution

We are given a particle that experiences a variable force F=2x. The work is done in displacing the body from x = x1x_1 to x = x2x_2. This work done can be found by integrating the force F with respect to x, from x = x1x_1 to x = x2x_2. Here, the force is a function of x, so we’ll assume it is acting along the x-axis.

Formula used:
W=∫x2x1FdxW = \int\limits_{{x_2}}^{{x_1}} {Fdx}

Complete step-by-step answer:
In the question, they’ve given a particle that is moving under the influence of force, F = 2x. Initially, the particle’s position is given by x = x1x_1 and the final position is given by x = x2x_2.
The work done by any force in displacing a body is given by
W=∫FdxW = \int {Fdx}
Where,
W is the work done
F is the force acting on the particle
As the position is changing from x1x_1 to x2x_2, we can integrate it from its initial position to the final position. The work done will be given by
W=∫x2x1FdxW = \int\limits_{{x_2}}^{{x_1}} {Fdx}
Substituting, F = 2x in the formula, we have
\eqalign{ & W = \int\limits_{{x_2}}^{{x_1}} {Fdx} \cr & \Rightarrow W = \int\limits_{{x_2}}^{{x_1}} {2xdx} \cr & \Rightarrow W = 2\int\limits_{{x_2}}^{{x_1}} {xdx} \cr & \Rightarrow W = 2\left| {\dfrac{{{x^2}}}{2}} \right|_{{x_2}}^{{x_1}} \cr & \Rightarrow W = 2 \times \dfrac{{{x_2}^2 - {x_1}^2}}{2} \cr & \Rightarrow W = {x_2}^2 - {x_1}^2 \cr}
Therefore, the work done in moving the particle is x22−x12{x_2}^2 - {x_1}^2.

So, the correct answer is “Option B”.

Note: Work done is actually the integral of the dot product of force and displacement. The dot product is the product of magnitudes and the cosine of the angle between them. In the question, we have not considered the dot product as the force and the displacement are in the same direction. Thus, the angle between them is zero and the cosine of zero is 1.