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 at t ge 3 second. Using impulse equation `int_0^1Fdt-(P_F-P_i)
int_3^5(40t-120)dt+int_5^(10)(200-120)dt=20v_1`
`40/2(25-9)-120(2)+80xx(10-5)=20v_1`
`320 - 240 + 400 = 20 v_1`
`therefore v_1=24=8v`
thereforev=3`