The Question is from neet-physics work-energy-power work-done-by-variable-force, and it was created by the author P.NARASIMHA

Question

A particle moves along the x-axis under the influence of a variable force given by $F(x) = \frac{a}{x^2} + bx$, where 'a' and 'b' are positive constants.  If the particle starts from rest at $x = x_0$, what is the work done by the force when the particle reaches $x = 2x_0$?

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: Option A
## CONCEPT FRAMEWORK
**Primary Concept**: Work-Energy Theorem
**Related Topics**: Variable Force, Integration
**Prerequisites**: Definition of work done by a variable force, basic integration
## SOLUTION PATHWAY
**Given Information**:
  * Force: $F(x) = \frac{a}{x^2} + bx$
  * Initial position: $x_i = x_0$
  * Final position: $x_f = 2x_0$
  * Initial velocity: $v_i = 0$ (starts from rest)
**Method & Steps**:
  1.  The work done by a variable force is given by the integral of the force with respect to displacement.
  Reasoning: The work done is the area under the force-displacement curve.
  Formula/Rule: $W = \int_{x_i}^{x_f} F(x) dx$
  2.  Substitute the given force and limits of integration into the formula.
  Working: $W = \int_{x_0}^{2x_0} (\frac{a}{x^2} + bx) dx$
  Key Point: Integrate each term separately.
  3. Evaluate the integral.
  Working: $W = [\frac{-a}{x} + \frac{bx^2}{2}]_{x_0}^{2x_0}$
  $W = [\frac{-a}{2x_0} + \frac{b(2x_0)^2}{2}] - [\frac{-a}{x_0} + \frac{bx_0^2}{2}]$
  $W = \frac{-a}{2x_0} + 2bx_0^2 + \frac{a}{x_0} - \frac{bx_0^2}{2}$
  $W = a(\frac{1}{x_0} - \frac{1}{2x_0}) + 2bx_0^2 - \frac{bx_0^2}{2}$
  $W = a(\frac{1}{x_0} - \frac{1}{2x_0}) + \frac{3bx_0^2}{2}$
  Result: The work done is $a(\frac{1}{x_0} - \frac{1}{2x_0}) + \frac{3bx_0^2}{2}$
  Verification: The result matches Option A.
## OPTION ANALYSIS
[A] ✓: This is the correct expression for the work done, as derived above.
[B] :x: The term involving 'a' has the sign reversed and the sign of b is correct.
[C] :x: The term involving 'b' is incorrect.
[D] :x: Both the terms involving 'a' and 'b' are incorrect.
## LEARNING AIDS
**Common Mistakes**:
  1. Incorrect integration of $\frac{1}{x^2}$. Remember that $\int x^{-2}dx = -x^{-1} + C$.
  How to avoid: Be careful with the power rule of integration.
  2. Incorrect substitution of limits. Make sure you subtract the lower limit result from the upper limit result.
  How to avoid: Use brackets to prevent sign errors.
**Quick Tips**:
  1. Remember that work done by a variable force is given by the integral of force with respect to displacement.
  2. Break the integral into simpler terms to avoid errors.

P.NARASIMHA · Answer

The correct Option is "$a(\frac{1}{x_0} - \frac{1}{2x_0}) + \frac{3bx_0^2}{2}$"

P.NARASIMHA · Answer

$a(\frac{1}{2x_0} - \frac{1}{x_0}) + \frac{3bx_0^2}{2}$

P.NARASIMHA · Answer

$a(\frac{1}{x_0} - \frac{1}{2x_0}) + \frac{bx_0^2}{2}$

P.NARASIMHA · Answer

$a(\frac{1}{2x_0} - \frac{1}{x_0}) + \frac{bx_0^2}{2}$

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