Question

in the arrangement shown in the figure the acceleration of block m1 and m2 is​

Answer 1

Given :

▪ A system of two blocks of masses m1 and m2 has been provided.

To Find :

▪ Acceleration of both blocks.

Concept :

✏ First, we have to find out relation b/w acceleration of both blocks.

✏ The best way to derive this relation is to apply concept of work done by tension force.

$\bigstar\sf\red{Work\:done\:by\:tension\:force=0}$

Constrained motion Equation :

$\rightarrow\sf\:\sum{T\:x=0}\\ \\ \rightarrow\sf\:\sum{T\:v=0}\\ \\ \rightarrow\sf\:\sum{T\:a=0}\\ \\ \star\sf\:Net\:tension\:force\:on\:m_1=3T\\ \\ \star\sf\:Net\:tension\:force\:on\:m_2=T\\ \\ \mapsto\sf\:T_1\:a_1+T_2\:a_2=0\\ \\ \mapsto\sf\:(3T)(-a_1)+(T)(a_2)=0\\ \\ \mapsto\underline{\boxed{\bf{\purple{3a_1=a_2}}}}$

Calculation :

⚾ Net force on block of mass m1 :

$\dashrightarrow\sf\:m_1g-3T=m_1a_1\\ \\ \dashrightarrow\sf\:3T=m_1g-m_1a_1\\ \\ \dashrightarrow\bf\:T=\dfrac{m_1}{3}(g-a_1)\rightarrow\:(1)$

⚾ Net force on block of mass m2 :

$\dashrightarrow\sf\:T-m_2g=m_2a_2\\ \\ \dashrightarrow\bf\:T=m_2(g+a_2)\rightarrow\:(2)$

Equating both equations, we get

$\Rightarrow\sf\:\dfrac{m_1}{3}(g-a_1)=m_2(g+a_2)\\ \\ \Rightarrow\sf\:\dfrac{m_1g}{3}-\dfrac{m_1a_1}{3}=m_2g+3m_2a_1\\ \\ \Rightarrow\sf\:m_1g-3m_2g=a_1(9m_2+m_1)\\ \\ \Rightarrow\underline{\boxed{\bf{\pink{a_1=\dfrac{m_1g-3m_2g}{m_1+9m_2}}}}}\\ \\ \implies\sf\:a_2=3a_1\\ \\ \implies\underline{\boxed{\bf{\orange{a_2=\dfrac{3(m_1g-3m_2g)}{m_1+9m_2}}}}}$

Answer 2

TO FIND:

• Acceleration of block m1 & m2

Work done by tension force = 0

Constrained equation of motion

→ ΣT_x = 0

→ ΣT_v = 0

→ ΣT_a = 0

♦ Net tension on block m1 = 3T

♦ Net tension on block m2 = T

Let

• Tension on block m1 = 3T be T_1
• Tension on block m2 = T be T_2

Now,

→ (T_1)(a_1) + (T_2)(a_2) = 1

→ 3a_1 = a_2

Net force on block of mass m1:

(m_1)g - 3T = m_1 × a_1

Finding T

[(m_1)(a_1)] - (m_1)g = 3T

Taking common

(m_1)(a_1 - g) = 3T

T = [(m_1)(a_1 - g)/3]

Assuming as equation 1

___________________________

Net force on block of mass m2:

T - (m_2)g = (m_2)(a_2)

T = (m_2)(g + a_2)

Assuming as equation 2

Equating the equations

→ [(m_1)(a_1 - g)/3] = (m_2)(g + a_2)

→ (m_1)(a_1 - g)/3 = [m_2(g) + 3(m_2)(a_1)]

→ (m_1)g - 3m_2(g) = a_1(9m_1 + m_2)

→ a_1 = (m_1)g - 3m_2(g)/(9m_1 + m_2)

→ a_2 = 3a_1

Substituting value of a

→ a_2 = 3[(m_1)g - 3m_2(g)]/(9m_1 + m_2)