A 20 kg block is originally at rest on a horizontal surface for which the coefficient of friction is `0.6` . A horizontal force F is applied such that it varies with time as shown in the figure (a) and (b) . If the speed of the block after 10 s is 8 v then find v . (Take g = `10 m// s^(2)`)

Answer

3

Explanation

`f_("limit") = 0.6 N` = 0.6 mg = `0.6 xx 20 xx 10 = 120 N`
The equation for the applied force is
F = 40 t = 200 , `" " ` for `t le 5 , " " 5 le t le 10`
The limiting friction acts as `t ge 3` second.
Using impulse equation ,
`overset(t) underset(0)(int) F "dt" = (p_(f) - p_(i))`
`overset(5)underset(3)(int)(40 t - 120) "dt" + overset(10)underset(5)(int) (200 - 120) "dt" = 20 v_(1)`
`(40)/(2) (25-9) - 120 (2) + 80 xx (10- 5) = 20 v_(1)`
`320 - 240 + 400 = 20 v_(1)`
`therefore " " v _(1) = 24 = 8v`
`therefore " " v = 3`