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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? |
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Answer» Solution :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 )` 
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