Question

At the moment $t=0$ the force $F=at$ is applied to a small body of mass $m$ resting on a smooth horizontal plane (a is a constant). The permanent direction of this forms an angle $\alpha$ with the horizontal. Find:
(a) The velocity of the body at the moment of its breaking off the plane;
(b) The distance traversed by the body up to this moment.

Answer 1

$\to$ let small body of mass m and indicate x-axis along

the horizontal plane and y-axis, perpendicular to it.

Let block breaks of the plate at t = to i.e. N- 0.

So, $N=mg-ato$ $sin\alpha =0$

or $t_0=\frac{mg}{asin\alpha }...\left(1\right)$

From in =max, for the body under integration

$m\frac{d{V}_x}{dt}=atcos\alpha$ ; Integrating within the limit for v/t

$m{\int }_0^{v}dvx=acos\alpha {\int }_{0}tdt$ (using eq (1))

So, $v=\frac{ds}{dt}=\frac{acos\alpha }{2m}{t}^2...\left(2\right)$

Integrating ,$E{q}^n$ (2) for s(t)

$s=\frac{acos\alpha }{2m}\frac{t^3}{3}...\left(3\right)$

using the values of $t=t_0$

from Eq (1) into Eqs (2) and (3)

$v=\frac{m{g}^{2}cos\alpha }{2asin\alpha }$ and $s=\frac{m^2{g}^{2}cos\alpha }{6a^{2}si{n}^3\alpha }$