A body of mass m is at rest on a smooth horizontal plane. A force F = kt is applied on the body making an angle `alpha` with the horizontal. In the equation of force, t is time and k is a constant. At the instant the object loses contact with the plane, how far will it move along the plane and what will be its velocity?

Explanation

As F = kt, force increases with time. Vertical component of the force = Fsin`alpha`. The body loses contact with the plane when the vertical force equals the weight of the body. i.e., Fsin`alpha` = mg . Let the time after which the body loses contact be `t_(0)`.
`therefore" " Fsin alpha = kt_(0) sin alpha = "mg"`
`therefore" "t_(0) = (mg)/(k sin alpha)`
Horizontal component of the force F = Fcos`alpha` = kt cos`alpha`
Hence horizontal acceleration = `("ktcos"alpha)/(m) = (dv_(x))/(dt)`
`therefore" " dv_(x) = ("ktcos"alpha)/(m)cdot dt `
Integrating , `v_(x) = ("kcos"alpha)/(m) (t^(2))/(2) + A `
`" "` [where A = integration constant]
As the body starts from rest, at t = 0, v = 0.
Inserting this in equation (1), A = 0.
`therefore" " v_(x) = (kcos alpha)/(2m) t^(2) `
If the displacement along the plane is s, then
`therefore" "v_(x) = (ds)/(dt) = ("kcos"alpha)/(2m) t^(2) or, ds = ("kcos"alpha)/(2m)t^(2)dt `
Integrating again, s = `("kcos"alpha)/(6m) t^(3) + B`
`" "` [where B = integration constant ]
As at the initial moment, the displacement is zero, i.e., at t = 0, s = 0.
From equation (3), we get B = 0.
Inserting the value of B in equation (3),
`s = ("kcos"alpha)/(6m)t^(3)`
Hence, displacement along the plane in time `t_(0)`,
`s_(0) = ("kcos"alpha)/(6m) t_(0)^(3) = ("kcos"alpha)/(6m) ((mg)/("ksin"alpha))^(3)`
or, `s_(0) = (m^(2)g^(3)cos alpha)/( 6k^(2)sin^(3) alpha) `
and corresponding velocity, `v_(0) = ("kcos"alpha)/(2m) ((mg)/("ksin"alpha))^(2)`
`therefore" " v_(0) = (mg^(2) cos alpha)/(2"k"sin^(2) alpha )`