Question

An object starts from rest and is acted upon by a variable force $F$ as shown in the 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. $\frac{2F_0}{\tan \theta }$
• B. $\frac{F_0}{\sin \theta }$
• C. $\frac{2F_0}{\cot \theta }$
• D. $\frac{F_0}{2\cos \theta }$

Answer 1

The correct option is A $\frac{2F_0}{\tan \theta }$

Here in figure, we can see that force is function of x , graph between force and displacement is straight line with intersects at -Fo on y-axis.

so, $F_{net}=F\left(x\right)$

we know from Newton's law of motion##,$

$F_{net}=ma=mv\frac{dv}{dt}$

so$,mv\frac{dv}{dx}=F\left(x\right)$

but $butF\left(x\right)=\left(\tan \theta \right)x-F_0$

$\Rightarrow m\underset{0}{\overset{v}{\int }}vdv=\underset{0}{\overset{x}{\int }}\left\{\left(\tan \theta \right)-F_0\right\}dx$

$\Rightarrow m\frac{v^2}{2}=\left(\tan \theta \right)\frac{x^2}{2}-F_{0}x$

for again come to rest $,v=0$

then$,\tan \theta \frac{x^2}{2}-F_{0}x=0$

$x=\frac{2F_0}{\tan \theta }$

Hence,

option $\left(A\right)$ is correct answer.