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(A) Given the force $F = -\alpha x^{2}$.According to Newton's second law,$ma = -\alpha x^{2}$,so $a = -\frac{\alpha}{m} x^{2}$.We know that $a = v \fr

Vedclass · Accepted Answer

$\left(\frac{3 m v_{0}^{2}}{2 \alpha}\right)^{\frac{1}{3}}$

Vedclass · Answer

(A) Given the force $F = -\alpha x^{2}$.According to Newton's second law,$ma = -\alpha x^{2}$,so $a = -\frac{\alpha}{m} x^{2}$.We know that $a = v \frac{dv}{dx}$.Substituting this,we get $v \frac{dv}{dx} = -\frac{\alpha}{m} x^{2}$.Separating variables,we have $v dv = -\frac{\alpha}{m} x^{2} dx$.Integrating both sides with limits from $v = v_{0}$ to $v = 0$ and $x = 0$ to $x = x_{f}$:$\int_{v_{0}}^{0} v dv = -\frac{\alpha}{m} \int_{0}^{x_{f}} x^{2} dx$.Evaluating the integrals:$\left[ \frac{v^{2}}{2} \right]_{v_{0}}^{0} = -\frac{\alpha}{m} \left[ \frac{x^{3}}{3} \right]_{0}^{x_{f}}$.$0 - \frac{v_{0}^{2}}{2} = -\frac{\alpha}{m} \frac{x_{f}^{3}}{3}$.$\frac{v_{0}^{2}}{2} = \frac{\alpha x_{f}^{3}}{3m}$.Solving for $x_{f}$:$x_{f}^{3} = \frac{3 m v_{0}^{2}}{2 \alpha}$.$x_{f} = \left( \frac{3 m v_{0}^{2}}{2 \alpha} \right)^{\frac{1}{3}}$.

$A$ particle is projected with velocity $v_{0}$ along the $x-$axis. $A$ damping force is acting on the particle which is proportional to the square of the distance from the origin,i.e.,$ma = -\alpha x^{2}$. The distance at which the particle stops is:

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