Answer:
The mass of the another block is 60 kg.
Explanation:
Given that,
Mass of block M= 100 kg
Height = 1.0 m
Time = 0.90 s
Let the mass of the other block is m.
We need to calculate the acceleration of each block
Using equation of motion
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
Put the value into the formula
[tex]1.0=0+\dfrac{1}{2}\times a\times(0.90)^2[/tex]
[tex]a=\dfrac{2\times1.0}{(0.90)^2}[/tex]
[tex]a=2.46\ m/s^2[/tex]
We need to calculate the mass of the other block
Using newton's second law
The net force of the block M
[tex]Ma=Mg-T[/tex]
[tex]T=Mg-Ma[/tex]....(I)
The net force of the block m
[tex]ma=T-mg[/tex]
Put the value of T from equation (I)
[tex]ma=Mg-Ma-mg[/tex]
[tex]m(a+g)=M(g-a)[/tex]
[tex]m=\dfrac{M(g-a)}{(a+g)}[/tex]
Put the value into the formula
[tex]m=\dfrac{100(9.8-2.46)}{2.46+9.8}[/tex]
[tex]m=59.8\ \approx60\ kg[/tex]
Hence, The mass of the another block is 60 kg.
The value of the mass of the second block along the pulley is 296.9 kg.
The given parameters;
mass of the block, m = 100 kg
height of the block, h = 1 m
time of motion, t = 0.9 s
The acceleration of the blocks is calculated as follows;
h = ut + ¹/₂at²
h = 0 + ¹/₂at²
h = ¹/₂at²
at² = 2h
[tex]a = \frac{2h}{t^2} \\\\a = \frac{2\times 1}{0.9^2} \\\\a = 2.469 \ m/s^2[/tex]
The acceleration of the two blocks across the pulley is given as;
[tex]a = \frac{m_2g}{m_1 + m_2} \\\\m_1 + m_2 = \frac{m_2g}{a} \\\\m_1 = \frac{m_2g}{a} - m_2\\\\m_1 = \frac{100 \times 9.8}{2.469} - 100\\\\m_1 = 396.9 - 100\\\\m_1 = 296.9 \ kg[/tex]
Thus, the mass of the other block is 296.9 kg.
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