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Byju's Answer

Standard XII

Physics

Principle of Superposition

An object of ...

Question

An object of mass $m$ has a speed $v_{0}$ as it passes through the origin, on it's way out along the $+ x$ axis, it is subjected to a retarding force given by $F_{x} = - A x$ . Here, $A$ is a positive constant. Find its $x -$ coordinate when it stops.

v_{0} \sqrt{\frac{m}{A}}

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2 v_{0} \sqrt{\frac{m}{A}}

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v_{0} \sqrt{\frac{2 m}{A}}

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v_{0} \sqrt{\frac{m}{2 A}}

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Solution

The correct option is A $v_{0} \sqrt{\frac{m}{A}}$
Using work energy theorem

$K_{f} - K_{i} = W_{F} = \int F d x$
$0 - \frac{1}{2} m v_{0}^{2} = \int_{0}^{x} - A x d x$
$∴ \frac{1}{2} m v_{0}^{2} = \frac{A x^{2}}{2}$
or $x = v_{0} \sqrt{\frac{m}{A}}$

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An object of mass m has a speed v0 as it passes through the origin, on it's way out along the +x axis, it is subjected to a retarding force given by Fx=−Ax. Here, A is a positive constant. Find its x− coordinate when it stops.

The correct option is A v0√mAUsing work energy theoremKf−Ki=WF=∫Fdx0−12mv20=∫x0−Axdx∴12mv20=Ax22or x=v0√mA

An object of mass $m$ has a speed $v_{0}$ as it passes through the origin, on it's way out along the $+ x$ axis, it is subjected to a retarding force given by $F_{x} = - A x$ . Here, $A$ is a positive constant. Find its $x -$ coordinate when it stops.

The correct option is A $v_{0} \sqrt{\frac{m}{A}}$
Using work energy theorem

$K_{f} - K_{i} = W_{F} = \int F d x$
$0 - \frac{1}{2} m v_{0}^{2} = \int_{0}^{x} - A x d x$
$∴ \frac{1}{2} m v_{0}^{2} = \frac{A x^{2}}{2}$
or $x = v_{0} \sqrt{\frac{m}{A}}$