Respuesta :

Given:

the two blocks have mass

[tex]m_A[/tex]

and

[tex]m_B[/tex]

friction is zero here.

Required:

the magnitude of the acceleration

Explanation:

it is given that the mass of the block mA is much greater than the mass of the mB.

we assume that mass of the block mB is slightly larger than the mA

look at the free body diagram

from the above figure, we can write the equation as

for block mB

[tex]m_Bg-T=m_Ba...(1)[/tex]

and for block mA

[tex]T=m_Aa.....(2)[/tex]

from the equation (1) and (2) we can write

[tex]a=\frac{m_Bg}{m_A+m_B}[/tex]

this is the acceleration of both blocks.

now assume that mass of mA is much greater than the mass mB

then from the above equation we can say that the acceleration is very less.

takes some real values,

we assume that mA= 35 kg and mB=1 kg

so the acceleration is

[tex]\begin{gathered} a=\frac{1\times9.8\text{ m/s}^2}{1+25} \\ a=0.37\text{ m/s}^2 \end{gathered}[/tex]

we can see that acceleration is very small.

as we increase the mass of the mA then acceleration of both is getting less .

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