MHB What are the Variables Needed to Solve a Pulley System Problem?

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To solve the pulley system problem involving two blocks connected by a massless rope, the forces acting on each block must be analyzed. For the block on the table (m1), the tension (T) equals the mass times acceleration (m1a), while for the hanging mass (m2), the equation accounts for gravitational force minus tension (m2g - T = m2a). By substituting T from the first equation into the second, the acceleration (a) can be derived as a = (m2g) / (m1 + m2), and the tension can be calculated as T = (m1m2g) / (m1 + m2). To find the final speed (vf) when the hanging mass reaches the floor, the kinematic equation can be applied, yielding vf = sqrt((2m2gΔy) / (m1 + m2) + vi^2). This systematic approach allows for the determination of acceleration, tension, and final speed in the pulley system.
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Two blocks are connected by a massless rope as shown below. The mass of the block on the table is 5.4 kg and the hanging mass is 2.1 kg. The table and the pulley are frictionless.
6-1-p-043.png
I need to find acceleration, the tension of the rope, and the speed when mass 2 hits the floor when it starts from rest and is initially located 1.3 m from the floor.
I need some help with the setup.
 
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Let's look at the forces on each object in turn. For $m_1$, we have:

$$\sum F_x=T=m_1a$$

And for $m_2$, we have:

$$\sum F_y=m_2g-T=m_2a$$

Now we have two equations in two unknowns...can you proceed?

To answer the second part of the question, I would use the kinematic equation:

$$\Delta y=\frac{v_f^2-v_i^2}{2a}$$

Solve that for $v_f$, and then use the given values (and the acceleration $a$ you found in the first part) to get your answer. :)
 
As a follow-up, we have when substituting for $T$ from the first equation into the second:

$$m_2g-m_1a=m_2a$$

Solving for $a$, we obtain:

$$a=\frac{m_2g}{m_1+m_2}$$

And so:

$$T=\frac{m_1m_2g}{m_1+m_2}$$

And finally:

$$v_f=\sqrt{\frac{2m_2g\Delta y}{m_1+m_2} + v_i^2}$$
 
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