Answer
(a) The acceleration of the objects is $0.6m/s^2$
(b) The tension in the left rope is $104N$ and that in the right rope is $230N$
Work Step by Step
We will analyze the forces that affect the motion of each object. We assume that this system will move towards the right, in the direction of the falling of the 25-kg object.
All objects also move at the same acceleration $a$.
1) Object 1 $(m_1=10kg)$
- The weight of object 1 $m_1g=10\times9.8=98N$
- The tension in rope 1 $T_1$
Since this object is pulled up with acceleration $a$, $$T_1-m_1g=m_1a$$ $$T_1-98=10a$$ $$T_1=10a+98 (1)$$
2) Object 3 $(m_2=25kg)$
- The weight of object 3 $m_3g=25\times9.8=245N$
- The tension in rope 2 $T_2$
Since this object goes down with acceleration $a$, $$m_3g-T_2=m_3a$$ $$245-T_2=25a$$ $$T_2=245-25a (2)$$
3) Object 2 $(m_2=80kg)$
- The tension in rope 1 $T_1$, which points leftward
- The tension in rope 2 $T_2$, which points rightward
- Kinetic friction $f_k$; since we assume the system moves towards the right, $f_k$ points leftward.
$$f_k=\mu_kF_N=\mu_km_2g=0.1\times80\times9.8=78.4N$$
Because this object moves rightward with acceleration $a$, $$T_2-T_1-f_k=m_2a$$ $$T_2-T_1-78.4=80a (3)$$
Replace $T_1$ from (1) and $T_2$ from (2) into (3): $$245-25a-10a-98-78.4=80a$$ $$115a=68.6$$ $$a=0.6m/s^2$$
From there, we have $T_1=104N$ and $T_2=230N$
