At `t = 0`, a force `F = at^(2)` is applied to s small body of mass m at an angle `alpha` resting on a smooth horizontal plane.

Options

A. Velocity of the body at the moment it breaks off the plane is `sqrt((mg^(3))/(9 a tan^(2) alpha sin alpha))`
B. The distance travelled by the body before breaking off the plane is `(mg^(2))/(12a sin alpha tan alpha)`
C. Its acceleration at the time of breaking off the plane is `g cot alpha`.
D. Time at which it breaks off the plane is `sqrt((mg)/(a sin alpha))`

Answer

A, B, C, D

Explanation

For breaking off the plane
`F sin alpha = mg`
`rArr a t_(0)^(2) sin alpha = mg " " rArr t_(0) = sqrt((mg)/(a sin alpha))`
Speed at time of breaking off,
`v = underset(0)overset(t_(0))int (at^(2) cos alpha)/(m) dt = (at_(0)^(2) cos alpha)/(3 m)`
`= (a cos alpha)/(3m) . (mg)/(a sin alpha) sqrt((mg)/(a sin alpha))`
`sqrt((mg^(3))/(9a tan^(2) alpha sin alpha))`
`= (F cos alpha)/(m) = (at_(0)^(2) cos alpha)/(m) = (amg)/(a sin alpha) . (cos alpha)/(m) = g`
`cot alpha`
`s = int v dt = underset(0)overset(t_(0))int (at^(3))/(3m) cos alpha dt`
`= (a)/(12m) t_(0)^(2) = (a)/(12m) (m^(2) g^(2) cos alpha)/(a^(2) sin^(2) alpha)`
`= (mg^(2))/(12a tan alpha sin alpha)`