Respuesta :
Answer
given,
m₁ = 19 kg
m₂ = 2.6 kg
radius = 8.9 cm
vertical distance = 3.8 m
acceleration due to gravity = 9.8 m/s²
a) [tex]a = \dfrac{F}{m}[/tex]
[tex]a = \dfrac{g\times (m_1-m_2)}{m_1+m_2}[/tex]
[tex]a = \dfrac{9.8\times (19-2.6)}{19.6+2.6}[/tex]
a = 7.24 m/s²
b) T = 2.6 × (9.8 + 7.24)
T = 44.30 N
A light, inextensible cord passes over a light, frictionless pulley with a radius of 8.9 cm. It has a 19 kg mass on the left and a 2.6 kg mass on the right, both hanging freely. The rate at which the two masses are accelerating when they pass each other is 7.44 m/s². The tension in the cord is 44.84 N.
In a light inextensible cord;
- The mass on the right side m₁ = 19 kg
- The mass on the left side m₂ = 2.6 kg
- The radius of the pulley r = 0.089 m
Using Newton's Law to the mass m₁
m₁ g - T = m₁ a
where;
- the tension T = m₁ (g -a) ----- Let this be equation (1)
For mass m₂,
- T - m₂ g = m₂a ------- Let this be equation (2)
equation both equation (1) and (2) together;
[tex]\mathbf{m_1(g-a) -m_2g = m_2a}[/tex]
[tex]\mathbf{m_1g-m_1a -m_2g = m_2a}[/tex]
By rearrangement and collecting like terms;
a(m₂ + m₁) = g(m₁ - m₂)
Making acceleration (a) the subject of the formula:
[tex]\mathbf{a = \dfrac{g(m_1-m_2)}{(m_2+m_1)}}[/tex]
[tex]\mathbf{a = \dfrac{9.8(19-2.6)}{(2.6+19)}}[/tex]
[tex]\mathbf{a = \dfrac{9.8(16.4)}{(21.6)}}[/tex]
a = 7.44 m/s²
From equation (1), the tension in the cord is:
Tension (T) = m₁ (g -a)
T = 19(9.8 - 7.44)
T = 19(2.36)
T = 44.84 N
Therefore, we can conclude that the rate at which the two masses accelerate when they pass each other is 7.44 m/s², and tension in the cord when they pass each other is 44.84 N
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