Question

A bar of mass $m$ resting on a smooth horizontal plane starts moving due to the force $F=mg/3$ of constant magnitude. In the process of its rectilinear motion the angle $\alpha$ between the direction of this force and the horizontal varies as $\alpha =$as, where $a$ is a constant, and $s$ is the distance traversed by the bar from its initial position. Find the velocity of the bar as a function of the angle $\alpha$.

Answer 1

Newtons second law of motion in projection form along with horizontal or x-axis

i.e $F_x=M{w}_x$ gives-

$F\cos \left(as\right)=mv\frac{d_v}{d_s}$

f$\cos \left(as\right)ds=mvdv$

Integrating over the limits for v(s)

$\frac{v}{m}{\int }_0^{\infty }\cos asds=\frac{v^2}{2}$

or

$v=\frac{\sqrt{2f\sin \alpha }}{ma}$

$v=\frac{\sqrt{2\frac{mg}{3}\sin \alpha }}{ma}$