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))

A 20 kg block is originally at rest on a horizontal surface for which

1.	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» Solution :`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`