A block of mass `m = 3 kg` is resting over a rough horizontal surface having coefficient of friction `mu = 1//3` The block is pulled to the right by appliying a force `F` inclined at angle `37^(@)` with the horizontal as shown in fig .The forceincreases with time according to law `F = 2r` newton Calculate its velocity `v` at `t = 10s (g = 10 ms^(-2))`

Answer

`(25)/(3)ms^(-1)`

Explanation

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At initial moment `(r = 0)` the force of `F = 2t = 0` it mean no force is acting on the block and the force F increases gradually with time Therefore the friction will come into existence and it will come time existence and it will keep the block in static equlibruim till limiting equilibrium is reached let the come to limiting equilibrium at `t = t_(0)` then FBD at that moment will be as shown in fig.
`N + 2l_(0) sin 37^(@) = mg`
For horizontal equilibrium.
`mu N = 2t_(0) cos 37^(@)` .
, Solving equation (i) and (ii)
`N = 24` Newton and `t_(0) = 5 sec`
At `t = t_(0)` block is in limiting equilibrium hence just after this moment it will start to slip over the surface threefour at `tgt t_(0)` the `FBD` of the block will be as shown in figure for vertical equilibrium

`N + 2t sin 37^(@)= mg`
In horizontal direction it will acceleration
`2t cos 37^(@) - mu N = ma`
From Eqs(iii) and (iv) `a = (1)/(3) (2t - 10)`
, But accfeleration ` a = dv//dt` Therefore
`dv = a,d = (1)/(3) (2t - 10)dt`
Integrating above equation with the condition with the conditions that at `t = 5s v = 0` and at `t = 10s t = ?`
`int_(0)^(1) dv = int_(5)^(10) (1)/(3) (2T - 10) .dt rArr v = (25)/(3)ms^(-1)`