At the moment t=0, a force F=kt (k is a constant) is applied to a small body of mass m resting on a smooth horizontal plane . The permanent direction of force F , makes an angle beta with the gorizontal . Calculate (i) The velocity of the body at the moment of its breaking off the plane . (ii) The distance traversed by the body upto this moment .

At the moment t=0, a force F=kt (k is a constant) is applied to a smal

1.	At the moment t=0, a force F=kt (k is a constant) is applied to a small body of mass m resting on a smooth horizontal plane . The permanent direction of force F , makes an angle beta with the gorizontal . Calculate (i) The velocity of the body at the moment of its breaking off the plane . (ii) The distance traversed by the body upto this moment .
Answer» Solution :The free BODY diagram of massm is shown in figure . For vertical equlibrium `F sin beta +N =mg "(or)" As F=kt` `:. Kt sin beta +N=mg`…..(1) EQUATION of MOTION of mass m is `F cos beta =ma` or `kt cos beta = m (dv)/(dt) (as a=(dv)/(dt)) , dv =(dt)/cos beta dt` Integrating `int_(0)^(v) dv =k/m cos beta int_(0)^(t) t dt ` `[v]_(0)^(v)=k/m cos beta [t^(2)/2]_(0)^(t) RARR v=(k cos beta)/(2m) t^(2)`....(2) At the moment of BREAKING off the plane N=0.if `tau` is the corresponding time , then ` `:. "From" (1) k tau sin beta =mg "or" tau=(mg)/(k sin beta)`....(3) `:. ` The velocity at the moment of breaking off the plane is given by putting `t=tau ` in equation (2) , so `(v=v_(0))` `v_(0)=(k cos beta)/(2m) tau^(2)=(k cos beta)/(2m)((mg)/(k sin beta))^(2)` `:. v_(0)=(mg^(2) cos beta)/(2k sin^(2) beta)` (b) Equation (2) may be expressed as `(dx)/(dt)=(k cos beta )/(2m) t^(2)` Integrating `[x]_(0)^(x)=(k cos beta)/(2m)[t^(3)/3]_(0)^(t)` `:. x=(k cos beta)/(6m)t^(3)` When the body breaks off the plane , we have `:. x_(0)=(k cos beta)/(6m) ((mg)/(k sin beta))^(3) =(m^(2)g^(3) cos beta )/(6k^(2) sin ^(3) beta)`