## Engineering Mechanics: Statics & Dynamics (14th Edition)

$t=2.11s$
The required time can be determined as follows: $\Sigma F_y=-mgsin60+F_N=0$ $\implies F_N=mgsin60$ and, $F_x=-mgcos60+F_f=ma_x$ $-mgcos60+mgsin60(\mu_s)=ma_x$ $\implies g(\mu_s sin60-cos60)=a_x$ This simplifies to: $a_x=1.89156m/s^2$ Now, $v_x=1.89156(t)-4m/s$ This simplifies to: $t=2.11s$