Answer :
Answer:
(a) The value of the ratio m₁/m₂ is 0.581
(b) the acceleration of the combined masses is 1.139 m/s²
Explanation:
Given;
The acceleration of force applied to M₁, a₁ = 3.10 m/s²
The same force applied to M₂ has acceleration, a₂ = 1.80 m/s²
Let this force = F
According Newton's second law of motion;
F = ma
(a) the value of the ratio m₁/m₂
since the applied force is same in both cases, M₁a₁ = M₂a₂
[tex]\frac{m_1}{m_2} = \frac{a_2}{a_1} \\\\\frac{m_1}{m_2} = \frac{1.8}{3.1} \\\\\frac{m_1}{m_2} = 0.581[/tex]
(b) the acceleration of m₁ and m₂ combined as one object under the action force F
F = ma
[tex]a = \frac{F}{M} \\\\a = \frac{F}{m_1 + m_2} \\\\a = \frac{F}{0.581m_2 + m_2}\\\\a = \frac{F}{1.581m_2}[/tex]
[tex]But, m_2 = \frac{F}{a_2} \\\\a = \frac{F}{1.581m_2} = \frac{F*a_2}{1.581F} \\\\a = \frac{a_2}{1.581} \\\\a = \frac{1.8}{1.581} = 1.139 \ m/s^2[/tex]
Therefore, the acceleration of the combined masses is 1.139 m/s²
a. The value of the ratio is 0.581.
b. The acceleration should be [tex]1.139 m/s^2[/tex]
Newton's second law of motion:
Since
The acceleration of force applied to [tex]M_1,a_1 = 3.10 m/s^2[/tex]
The same force applied to[tex]M_2[/tex] has acceleration, [tex]a_2 = 1.80 m/s^2[/tex]
Here we assume force be F
Now we know that
F = ma
a. The ratio should be
[tex]m_1\div m_2 = a_2\div a_1\\\\m_1\div m_2 = 1.8\div 3.1[/tex]
= 0.581
b. The acceleration should be
f = ma
a = F/m
[tex]a = f/m_1+m_2\\\\= F/0.581 + m_2\\\\=F/1.581m_2[/tex]
However [tex]m_2 = F/a_2[/tex]
So,
[tex]a = F\div 1.581m_2\\\\= f\times a_2/1.581F\\\\= a_2/1.581\\\\= 1.8/1.581\\\\= 1.139 m/s^2[/tex]
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