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`.

Explanation

Newton's second law of motion in projection form, along horizontal or x-axis i.e.
`F_x=mw_x` gives.
`Fcos (as)=mv(dv)/(ds)` (as `alpha=as`)
or, `Fcos(as)ds=mvdv`
Integrating, over the limits for `v(s)`
`F/m underset0overset(oo)int cos (as) ds=v^2/2`
or `v=sqrt((2Fsin alpha)/(ma))`
`=sqrt(2g sin alpha//3a)` (using `F=(mg)/(3)`)
which is the sought relationship.