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 origin i.e. ma = $-$ $\alpha$x2. The distance at which the particle stops :
Solution
Given, speed of projection = v<sub>0</sub><br/><br/>Damping force, F = ma = $-$ $\alpha$x<sup>2</sup><br/><br/>$\Rightarrow$ a = $-$ $\alpha$x<sup>2</sup> / m<br/><br/>Also, $a = v{{dv} \over {dx}}$<br/><br/>$\Rightarrow vdv = a\,dx = - {\alpha \over m}{x^2}dx$<br/><br/>Integrating both sides, we get<br/><br/>$\int_{{v_0}}^v {vdv = \int_0^x { - {\alpha \over m}{x^2}dx} }$<br/><br/>$$ \Rightarrow \left( {{{{v^2}} \over 2}} \right)_{{v_0}}^0 = - {\alpha \over m}\left( {{{{x^3}} \over 3}} \right)_0^x$$<br/><br/>$$ \Rightarrow 0 - v_0^2/2 = - {\alpha \over m}{{{x^3}} \over 3} \Rightarrow x = {\left( {{{3m} \over 2}{{v_0^2} \over \alpha }} \right)^{1/3}}$$
About this question
Subject: Physics · Chapter: Laws of Motion · Topic: Newton's First Law
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