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4-1.Newton's Laws of Motion
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An object starts from rest and is acted upon by a variable force $F$ as shown in figure. If $F_0$ is the initial value of the force, then the position of the object, where it again comes to rest will be
A$\frac{2 F_0}{\tan \alpha}$
B$\frac{F_0}{\sin \alpha}$
C$\frac{2 F_0}{\cot \alpha}$
D$\frac{F_0}{2 \cos \alpha}$
Solution
(a)
$F-x$ curve is straight line. Equation of $F$ in terms of $x$ can be written as
$F=x \tan \alpha-F_0$
$a=\frac{v d v}{d x}=\frac{F}{m}=\frac{x \tan \alpha}{m}-\frac{F_0}{m}$
Integrating both sides
$\frac{v^2-x^2}{2}=\frac{x^2 \tan \alpha}{2 m}-\frac{F_0 x}{m}=0$
$\frac{x \tan \alpha}{2}=F_0$
$x=\frac{2 F_0}{\tan \alpha}$
$F-x$ curve is straight line. Equation of $F$ in terms of $x$ can be written as
$F=x \tan \alpha-F_0$
$a=\frac{v d v}{d x}=\frac{F}{m}=\frac{x \tan \alpha}{m}-\frac{F_0}{m}$
Integrating both sides
$\frac{v^2-x^2}{2}=\frac{x^2 \tan \alpha}{2 m}-\frac{F_0 x}{m}=0$
$\frac{x \tan \alpha}{2}=F_0$
$x=\frac{2 F_0}{\tan \alpha}$
Standard 11
Physics
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