Question

A variable force of the form F = at is applied on a block of mass m
from time t = 0, kept on a smooth horizontal surface. Find the velocity
of the block at the instant it leaves the surface.
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Answer 1

Though Sorry for bad handwriting

Answer 2

Answer:

$v = a\dfrac{(mg)^{2}  }{2m(asin\theta )^{2} } cos\theta$

Explanation:

Since we know that the force has two component $Fcos\theta$ along horizontal and $Fsin\theta$ along vertical.

$Fsin\theta = mg$......................(1)

We have F =at

$atsin\theta = mg$

$t = \dfrac{mg}{asin\theta }$..............................(2)

$Fcos\theta = ma$

$atcos\theta = m\dfrac{dv}{dt}$

$atcos\theta dt = mdv$

integrate both sides we get

$\int\ {atcos\theta dt }  = \int\ mdv$

$a\dfrac{t^{2} }{2} cos\theta = mv$

So,

$v = a\dfrac{t^{2} }{2m} cos\theta$........................................(3)

Using (2) and (3) we get,

So, required velocity is,

$v = a\dfrac{(mg)^{2}  }{2m(asin\theta )^{2} } cos\theta$