3. determine the acceleration of the system and the tension of the rope.

Explanation:

Step1: Analyze forces on 16 - kg block

For the 16 - kg block on the inclined plane, the gravitational - force component along the plane is $F_{g1\parallel}=m_1g\sin\theta$ and the normal force is $N = m_1g\cos\theta$. The frictional force is $f=\mu N=\mu m_1g\cos\theta$, where $m_1 = 16$ kg, $g = 9.8$ m/s², $\theta = 25^{\circ}$, and $\mu=0.65$. $F_{g1\parallel}=(16)(9.8)\sin25^{\circ}\approx66.3$ N $N=(16)(9.8)\cos25^{\circ}\approx141.9$ N $f=(0.65)(16)(9.8)\cos25^{\circ}\approx92.2$ N

Step2: Analyze forces on 20 - kg block

The gravitational force on the 20 - kg block is $F_{g2}=m_2g=(20)(9.8) = 196$ N. Let the acceleration of the system be $a$ and the tension in the rope be $T$. For the 16 - kg block, using Newton's second law $T - F_{g1\parallel}-f=m_1a$. For the 20 - kg block, $F_{g2}-T=m_2a$. Adding these two equations: $F_{g2}-F_{g1\parallel}-f=(m_1 + m_2)a$. Substitute the values: $196-66.3 - 92.2=(16 + 20)a$. $37.5 = 36a$.

Step3: Solve for acceleration

$a=\frac{37.5}{36}\approx1.04$ m/s².

Step4: Solve for tension

Substitute $a$ into the equation $F_{g2}-T=m_2a$. $T=F_{g2}-m_2a$. $T = 196-(20)(1.04)=196 - 20.8 = 175.2$ N.

Answer:

The acceleration of the system is approximately $1.04$ m/s² and the tension of the rope is approximately $175.2$ N.