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

`x-`direction`:` `Fcostheta=ma`
`a=(F)/(m)costheta=(lamdat)/(m)costheta`
`(dv)/(dt)=(lamdacostheta)/(m)l`
`int_(0)^(v)dv=(lamdacostheta)/(m)int_(0)^(v)tdt`
`v=(lamdacosthetat^(2))/(2m)` (i)
`(ds)/(dt)=(lamdacosthetat^(2))/(2m)`
`int_(0)^(s)ds=(lamdacostheta)/(2m)int_(0)^(t)t^(2)dt`
`s=(lamdacosthetat^(2))/(6 m)` (ii)
`y-`direction`:` `N+Fsintheta=mg`
`N+lamdatsintheta=mg`
When the block loses contact with the ground. `N=0`
`t=t_(0)=(mg)/(lamdasintheta)`
`v=(lamdacosthetat_(0)^(2))/(2m)=(lamdacostheta)/(2m)((mg)/(lamdasintheta))^(2)=(mg^(2)costheta)/(2lamdasin^(2)theta)`
`s=(lamdacosthetat_(0)^(2))/(6m)=(lamdacostheta)/(6m)((mg)/(lamdasintheta))^(3)=(m^(2)g^(3)costheta)/(6lamda^(2)sin^(3)theta)`