Question

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. $\frac{2F_0}{\tan \alpha }$
• B. $\frac{F_0}{\sin \alpha }$
• C. $\frac{2F_0}{\cot \alpha }$
• D. $\frac{F_8}{2\cos \alpha }$

Answer 1

Answer: (A)

$F-x$ curve is straight line. Equation of $F$ in terms of $x$ can be written as
$F=x\tan \alpha -F_0$
$a=\frac{vdv}{dx}=\frac{F}{m}=\frac{x\text{ tan }\alpha }{m}-\frac{F_0}{m}$
Integrating both sides
$\frac{v^2-x^2}{2}=\frac{x^2\tan \alpha }{2m}-\frac{F_{0}x}{m}=0$
$\frac{x\tan \alpha }{2}=F_0$
$x=\frac{2F_0}{\tan \alpha }$