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

$a_A=9.66ft/s^2$, $a_B=15ft/s^2$
The acceleration of each block can be determined as follows: We know that $\Sigma F_y=0$ $\implies N-W=0$ $\implies N=W=8lb$ and $\Sigma F_x=ma_x$ $\implies kx-\mu_k N=m_A a_A$ We plug in the known values to obtain: $20\times 0.2-(0.2\times 8)=\frac{8}{32.2}a_A$ This simplifies to: $a_A=9.66ft/s^2$ Now for the second block $\Sigma F_y=0$ $\implies N-W=0$ $\implies N=W=6lb$ and $\Sigma F_x=ma_x$ $\implies kx-\mu_k N=m_B a_B$ We plug in the known values to obtain: $20\times 0.2-0.2\times 6=\frac{6}{32.2}a_B$ This simplifies to: $a_B=15ft/s^2$