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A bar of mass $m$ resting on a smooth horizontal plane starts moving due to a constant force $F$. In the process of its rectilinear motion the angle $\theta$ between the direction of this force and the horizontal varies as $\theta=k x$, where $k$ is a constant and $x$ is the distance traversed by the bar from its initial position. The velocity $(v)$ of the bar as a function of the angle $\theta$ is
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The correct answer is:
$v=\sqrt{\frac{2 F \sin \theta}{m k}}$
Component of force responsible for horizontal motion is $F \cos \theta$.
So, acceleration of bar, $a=\frac{F \cos \theta}{m}$
As acceleration, $a=\frac{d v}{d t}$, we have
$\Rightarrow \quad \frac{d v}{d t}=\frac{F \cos \theta}{m}$
we can also express this as,
$\begin{array}{ll}\Rightarrow & \frac{d v}{d x} \cdot \frac{d x}{d t}=\frac{F \cos \theta}{m} \\ \Rightarrow & v d v=\frac{F \cos \theta}{m} d x\end{array}$
Integrating both sides and putting $\theta=k x$, we get
$\int v d v=\int \frac{F \cos k x}{m} d x$
$\begin{array}{ll}\Rightarrow & \frac{v^2}{2}=\frac{F}{m} \int \cos k x \cdot d x \\ \Rightarrow & \frac{v^2}{2}=\frac{F}{m} \cdot\left(\frac{\sin k x}{k}\right) \\ \Rightarrow & v^2=\frac{2 F \sin k x}{m k} \\ \Rightarrow & v=\sqrt{\frac{2 F \sin \theta}{m k}}\end{array}$
So, acceleration of bar, $a=\frac{F \cos \theta}{m}$
As acceleration, $a=\frac{d v}{d t}$, we have
$\Rightarrow \quad \frac{d v}{d t}=\frac{F \cos \theta}{m}$
we can also express this as,
$\begin{array}{ll}\Rightarrow & \frac{d v}{d x} \cdot \frac{d x}{d t}=\frac{F \cos \theta}{m} \\ \Rightarrow & v d v=\frac{F \cos \theta}{m} d x\end{array}$
Integrating both sides and putting $\theta=k x$, we get
$\int v d v=\int \frac{F \cos k x}{m} d x$
$\begin{array}{ll}\Rightarrow & \frac{v^2}{2}=\frac{F}{m} \int \cos k x \cdot d x \\ \Rightarrow & \frac{v^2}{2}=\frac{F}{m} \cdot\left(\frac{\sin k x}{k}\right) \\ \Rightarrow & v^2=\frac{2 F \sin k x}{m k} \\ \Rightarrow & v=\sqrt{\frac{2 F \sin \theta}{m k}}\end{array}$
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