Question

A block of mass $m$ resting on a smooth horizontal plane starts moving due to a constant force $F=mg/3$ of constant magnitude. In the process of its rectilinear motion the angle $\theta$ varies as $\theta =\lambda s$, is constant and $s$ is the distance traversed by the block from its initial position. Find the velocity of the block as a function of the angle $\theta$ . Also show that the block will never lose contact with the ground.

Answer 1

Taking the $x-$component of force:

$F_x=F\cos \theta F\cos \left(\lambda s\right)ds=mv\frac{dv}{ds}F\cos \left(\lambda s\right)ds=mvdv$

Integrating both sides

$\frac{F}{m}\int \cos \left(\lambda s\right)ds=\frac{v^2}{2}\Rightarrow v=\sqrt{\frac{2F\sin \theta }{m\lambda }}=\sqrt{2g\sin \left(\theta /3\lambda \right)}$