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 constant).
The permanent direction of this force forms an angle `alpha` with the horizontal (figure). 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.

Explanation

First of all let us draw the free body diagram for the small body of mass m and indicate x-axis along the horizontal plane and y-axis, perpendicular to it, as shown in the figure.
Let the block breaks off the plane at `t=t_0` i.e. `N=0`
So, `N=mg-at_0sin alpha=0`
or, `t_0=(mg)/(a sin alpha)` (1)
From `F_x=mw_x`, for the body under investigation:
`md v//dt=at cos alpha`, Integrating within the
limits for `v(t)`
`m underset(0)oversetvint dv_x=a cos alpha underset0int t dt` (using Eq. 1)
So, `v=(ds)/(dt)=(a cos alpha)/(2m)t^2` (2)
Integrating, Eqs. (2) for s(t)
`s=(a cos alpha)/(2m)t^3/3` (3)
Using the value of `t=t_0` from Eq. (1), into Eqs. (2) and (3)
`v=(mg^2cos alpha)/(2 a sin^2alpha)` and `s=(m^2g^3cos alpha)/(6a^2sin^3alpha)`