Answer:
[tex]v_i =0[/tex] since the mass start at rest
[tex]v_f = 3.54 \ m/s[/tex]
Explanation:
Given that :
Two masses of 24 kg and 12 kg are suspended by a pulley
so; let [tex]m_1[/tex] = 24 kg and [tex]m_2[/tex] = 12 kg
radius r = 6.2 cm
mass M = 5.0 kg
The moment of inertia on the pulley is ;
[tex]I = \frac{1}{2}Mr^2[/tex]
For the force equation with mass m ; we have:
[tex]m_1g -T_1 = m_1a[/tex]
[tex]T_1 =m_1g-m_1a[/tex] -------- Equation (1)
Force equation of the mass [tex]m_2[/tex] Â can be written as:
[tex]T_2 -m_2g = m_2a[/tex]
[tex]T_2 = m_2a+m_2g[/tex] Â Â -------- Equation (2)
The torque on the pulley is expressed as:
[tex](T_1-T_2) r = I \alpha[/tex]
where [tex]a= r* \alpha[/tex]
Then;
[tex](T_1-T_2) r =\frac{1}{2}Mr^2 \frac{a}{r}[/tex]
[tex]T_1 -T_2 = \frac {Ma}{2}[/tex] ----- Equation (3)
Replacing equation (1) and (2) Â into equation (3) ; we have:
[tex](m_1g -m_1a) -(m_2g + m_2a) = \frac{Ma}{2}[/tex]
[tex](24*9.8 -24a) -(12*9.8 + 12a) = \frac{5a}{2}[/tex]
[tex]235.2 - 24a - 117.6-12a = \frac{5a}{2}[/tex]
[tex]117.6 - 36a= \frac{5a}{2}[/tex]
[tex]2(117.6-36a)= \frac{5a}{2}[/tex]
235.2 - 72a  = 5a
235.2 = 5a + 72a
235.2 = 77a
a = [tex]\frac{235.2}{77}[/tex]
a= 3.05 m/s²
Let's consider the mass [tex]m_1[/tex] of the motion, with initial velocity
[tex]v_i =0[/tex] since the mass start at rest
distance traveled (s) = 2.05 m
acceleration = 3.05 m/s²
Using the formula:
[tex]v_f^2 = v_i^2 + 2 as \\ \\ v_f^2 =0^2+2(3.05)*2.05 \\ \\ v_f^2 = 12.505 \\ \\ v^2_f = \sqrt{12.505} \\ \\ v_f = 3.54 \ m/s[/tex]
