Question

Please help with this question as soon as possible.​

Answer 1

Answer:

$\boxed{\mathfrak{(1) \ \dfrac{2F_0}{tan\alpha}}}$

Explanation:

Here force given is variable.

Hence, we apply method of integration for getting desired result.

Equation of force is in the form of y = mx + c

$\sf m = -tan \alpha \\ \sf c = F_0$

$\rm \implies F = -tan\alpha \: x + F_0$

$\sf F = ma \\ \\ \sf\sf F = m \frac{dv}{dt} \times \frac{dx}{dx} \\ \\ \sf F = m \frac{dx}{dt} . \frac{dv}{dx} \\ \\ \sf F = mv \frac{dv}{dx}$

$\rm \implies mv. \dfrac{dv}{dx} = -tan\alpha \: x + F_0$

Initial and final velocity both are zero and 'x' is final position of object when it comes to rest again.

So,

$\rm \implies \int\limits^0_0mv.dv = \int\limits^x_0(-tan\alpha \: x + F_0).dx \\ \\ \rm \implies 0 = ( - tan \alpha \dfrac{ {x}^{2} }{2} + F_0x)\Big| _0^x \\ \\ \rm \implies F_0x - tan \alpha \dfrac{ {x}^{2} }{2}= 0 \\ \\ \rm \implies x( F_0 - tan \alpha \dfrac{ x }{2})= 0 \\ \\ \\ \\ \rm \implies x = 0 \: \: \: \: or \: \: \: \: F_0 - tan \alpha \dfrac{ x }{2} = 0 \\ \\ \rm \implies tan \alpha \dfrac{ x }{2} = F_0\\ \\ \rm \implies x = \dfrac{2F_0}{tan \alpha }$

$\therefore$ Position of object where it again comes to rest = $\rm \dfrac{2F_0}{tan \alpha }$