Question

The acceleration of block of mass M is​

Answer 1

To Find :

▪ Acceleration of block of mass M.

Concept :

↗ This question is completely based on the concept of constrained motion.

↗ First of all we have to find out relation b/w acceleration of both blocks.

↗ The best way to find this type of relation is to apply concept of work done by tension force as we know that, net work done by tension force is zero.

Diagram :

Please, see the attachment for better understanding.

Calculation :

✴ Let acceleration of block of mass M be a and that for block of mass m be a'.

▪ Net tension force on block of mass M = T

▪ Net tension force on block of mass m_o = 5T

$\dashrightarrow\sf\:\red{\sum T\:{\tiny{\bullet}}\:a=0}\\ \\ \dashrightarrow\sf\:(T)(-a)+(5T)(a')=0\\ \\ \dashrightarrow\sf\:T\:a=5T\:a'\\ \\ \dashrightarrow\boxed{\bf{a=5a'}}\:\rightarrow\:(1)$

⭐ Force equation for block (M) :

$\Rightarrow\sf\:Mg-T=Ma\\ \\ \Rightarrow\bf\:a=\dfrac{Mg-T}{M}\:\rightarrow\:(2)$

⭐ Force equation for block (m_o) :

$\Rightarrow\sf\:5T-m_og=m_oa'\\ \\ \Rightarrow\bf\:a'=\dfrac{5T-m_og}{m_o}\:\rightarrow\:(3)$

Putting values in equation (1)

$\rightarrow\sf\:\dfrac{Mg-T}{M}=5\times\dfrac{5T-m_og}{m_o}\\ \\ \rightarrow\sf\:Mm_og-m_oT=25MT-5Mm_og\\ \\ \rightarrow\sf\:Mm_og+5Mm_og=25MT+m_oT\\ \\ \rightarrow\boxed{\bf{T=\dfrac{6Mm_og}{25M+m_o}}}$

Putting value of T in equation (2)

$\twoheadrightarrow\sf\:a=\dfrac{Mg-T}{M}\\ \\ \twoheadrightarrow\sf\:a=\dfrac{Mg-\frac{6Mm_og}{25M+m_o}}{M}\\ \\ \twoheadrightarrow\sf\:a=\dfrac{25Mg+m_og-6m_og}{25M+m_o}\\ \\ \twoheadrightarrow\underline{\boxed{\bf{\purple{a=\dfrac{5(5M-m_o)g}{25M+m_o}}}}}\:\orange{\bigstar}$

Option-A is correct !!

Answer 2

$\bigstar$ $\sf{\green{\underline{\underline{\orange{Correct\:Option\::(a)}}}}}$

$\bigstar$ $\sf{\green{\underline{\underline{\orange{CONCEPT:-}}}}}$

• When acceleration of block is asked in a pully system , then

1. First draw the free body diagram.
2. Second write the equation of force for the block .
3. Finally calculate the value of acceleration .

$\bigstar$ $\sf{\green{\underline{\underline{\orange{To\:Find:-}}}}}$

• The acceleration of block of mass “M” .

$\bigstar$ $\sf{\green{\underline{\underline{\orange{SOLUTION:-}}}}}$

• See the above diagram in pic, to clarify the concept of “Free body diagram” .

Let , block of mass “M” descending down with acceleration “a” and mass ${m_o}$ ascending up with acceleration “${a_1}$" .

☞ Equation for Block of mass “M”

• ${Mg\:-\:T\:=\:Ma}$ ----------(1)

☞ Equation for Block of mass

“${m_o}$” ,

✔️Here total tension is “5T” .

• ${5T\:-\:m_o\:g\:=\:m_o\:a_1}$ --------(2)

☞ Equation for Tension on both block,

• ${T(-a)\:+\:5T\:(a_1)\:=\:0}$
• $\tt{\implies\:T(a)\:=\:5T\:(a_1)}$
• $\tt{\implies\:a\:=\:5\:(a_1)}$ ----------(3)

☞ From the above equations we get,

• ${a\:=\:{\dfrac{Mg\:-\:T}{M}}}$ ---------(A)

• ${a_1\:=\:{\dfrac{5T\:-\:m_og}{m_o}}}$ ------(B)

☞ Now put the value of Equation (3) in Equation (B) and equate that with Equation (A) we get,

• ${T\:=\:{\dfrac{6\:M\:m_o\:g}{25M\:+\:m_o}}}$

☞ Now put the value of “T” in Equation (A) we get,

• ${a_M\:=\:{\dfrac{5(5M\:-\:m_o)g}{25M\:+\:m_o}}}$

$\bigstar\:\underline{\boxed{\bf{\red{Required\:Answer\::\:a_M\:=\:{\dfrac{5(5M\:-\:m_o)g}{25M\:+\:m_o}}}}}}\:\bigstar$