At t=0, a force F = kt is applied on a block of mass m making an angle `alpha` with the horizontal . Suppose surfaces to be smooth. Find the velocity of the body at the time of breaking off the plane.

Explanation

The free body diagram of mass m is shown in figure . For vertical equilibrium
` 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=(kt)/(m) cos beta dt `
Integrating ` int_0^(v) dv = (k)/(m) cos beta int_0^(t) dt `
`[v]_0^(v) = (k)/(m) cos beta [(t^(2))/(2)]_0^(t) implies v= (k cos beta )/( 2 m ) 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 )/( 2 m) tau^(2) = ( k cos beta )/(2 m) ((mg)/(k sin beta ))^(2)`
`:. v_0=(mg^(2) cos beta)/( 2 k 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 )/( 6 m) 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 )`