Question

*12. for the pulley system illustrated to the right, when the masses are released, what is the final speed of the 12 kg mass just before it hits the floor?
(image of pulley system with 4.0 kg and 12 kg masses, distance 5.0 m)

Explanation:

Step1: Define energy conservation

The total mechanical energy of the system is conserved. The potential energy lost by the 12 kg mass is converted into the kinetic energy of both masses and the potential energy gained by the 4.0 kg mass. Let $m_1=12\ \text{kg}$, $m_2=4.0\ \text{kg}$, $h=5.0\ \text{m}$, $g=9.8\ \text{m/s}^2$, and $v$ be the speed of both masses (they move at the same speed due to the pulley).
$$m_1gh = m_2gh + \frac{1}{2}(m_1+m_2)v^2$$

Step2: Rearrange for $v$

Isolate the kinetic energy term, then solve for $v$:
$$(m_1 - m_2)gh = \frac{1}{2}(m_1+m_2)v^2$$
$$v = \sqrt{\frac{2(m_1 - m_2)gh}{m_1+m_2}}$$

Step3: Substitute values

Plug in the given values into the formula:
$$v = \sqrt{\frac{2(12-4.0)\times9.8\times5.0}{12+4.0}}$$
$$v = \sqrt{\frac{2\times8.0\times9.8\times5.0}{16}}$$
$$v = \sqrt{\frac{784}{16}} = \sqrt{49}$$

Answer:

$7.0\ \text{m/s}$