The Question is from neet-physics work-energy-power work-energy-theorm-for-variable-force, and it was created by the author VENKATESWARARAO

Question

A particle moves along the x-axis under the influence of a variable force $F(x) = \frac{a}{x^2} - \frac{b}{x^3}$, 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 $x = 2x_0$?

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Work-Energy Theorem
	**Related Topics**: Integration, Variable Force, Potential Energy
	**Prerequisites**: Understanding of work done by a variable force, basic integration
## SOLUTION PATHWAY
**Given Information**:
	Force function: $F(x) = \frac{a}{x^2} - \frac{b}{x^3}$
	Initial position: $x_0$
	Final position: $2x_0$
	Initial velocity: 0 (starts from rest)
**Method & Steps**:
	1.	Calculate the work done by the variable force.
		Reasoning: Work done by a variable force is given by the integral of the force with respect to displacement.
		Formula/Rule: $W = \int_{x_1}^{x_2} F(x) dx$
	2.	Integrate the force function from $x_0$ to $2x_0$.
		Working: $W = \int_{x_0}^{2x_0} (\frac{a}{x^2} - \frac{b}{x^3}) dx = \int_{x_0}^{2x_0} (ax^{-2} - bx^{-3}) dx$
			$W = [-ax^{-1} + \frac{b}{2}x^{-2}]_{x_0}^{2x_0} = [-\frac{a}{x} + \frac{b}{2x^2}]_{x_0}^{2x_0}$
			$W = (-\frac{a}{2x_0} + \frac{b}{2(2x_0)^2}) - (-\frac{a}{x_0} + \frac{b}{2x_0^2})$
			$W = -\frac{a}{2x_0} + \frac{b}{8x_0^2} + \frac{a}{x_0} - \frac{b}{2x_0^2}$
			$W = \frac{a}{2x_0} - \frac{3b}{8x_0^2}$
		Key Point: Remember to evaluate the integral at both limits and subtract the lower limit from the upper limit.
	3.	State the final answer.
		Result: The work done is $W = \frac{a}{2x_0} - \frac{3b}{8x_0^2}$
		Verification: The derived expression matches option A.
 ## OPTION ANALYSIS
[A] ✓: The work done by the force is correctly calculated as $\frac{a}{2x_0} - \frac{3b}{8x_0^2}$.
[B] :x: This option does not match the correct calculation of the work done.
[C] :x: This option does not match the correct calculation of the work done.
[D] :x: This option does not match the correct calculation of the work done.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Incorrect integration of the force function. Always remember the power rule for integration and be careful with signs.
		2.[Error 2]: Forgetting to evaluate the integral at both the upper and lower limits. Ensure you subtract the value at the lower limit from the value at the upper limit.
	**Quick Tips**:
		1.[Helpful shortcut]: When dealing with work done by a variable force, always remember to integrate the force function with respect to the displacement.
		2.[Memory aid]: Remember the power rule for integration: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (where n ≠ -1).

VENKATESWARARAO · Answer

The correct Option is "a/(2x0) - 3b/(8x0²)"

VENKATESWARARAO · Answer

a/(x0) - b/(x0²)

VENKATESWARARAO · Answer

a/(4x0) - b/(2x0²)

VENKATESWARARAO · Answer

2a/(x0) - 3b/(x0²)

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