A particle is projected with velocity $$v0$$ along $$x-axis$$. A damping force is acting on the particle which is proportional to the square of the distance from the i.e. $$ma=$$−$$ax2$$. The distance at which the particle stops: [Take m $$=1kg$$]

Question

A particle is projected with velocity $$v0$$  along $$x-axis$$. A damping force is acting on the particle which is proportional to the square of the distance from the i.e. $$ma=$$−$$ax2$$. The distance at which the particle stops: [Take m $$=1kg$$]

Infinity Learn · Accepted Answer

Given : $$ma=$$−α$$x2$$⇒$$mvdvdx=$$−α$$x2$$⇒m∫$$v00$$ $$vdv=$$−α∫$$0x$$ $$x2dx$$  ⇒$$m0-v022=-$$α$$x33-0$$⇒$$x=3mv022$$α$$13$$⇒$$x=3v022$$α$$13$$        [put m $$=1kg$$]

A particle is projected with velocity $v_{0}$ along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the i.e. $m a = - a x^{2} .$ The distance at which the particle stops: [Take m =1kg]

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A particle is projected with velocity v0 along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the i.e. ma=−ax2. The distance at which the particle stops: [Take m =1kg]

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A particle is projected with velocity $v_{0}$ along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the i.e. $m a = - a x^{2} .$ The distance at which the particle stops: [Take m =1kg]