Question

In the given figure ,$m_{1} = \: 1kg ,\: m_{2} = 4 kg \: , \: m_{3} = 1 kg$ ...…..... refer to attachment

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Answer 1

Solution :

⏭ Given:

✏ A pulley system is provided.

• mass of m1 = 1kg
• mass of m2 = 4kg
• mass of m3 = 1kg

⏭ To Find:

✏ Acceleration of all three blocks

⏭ Assumptions:

✏ All the strings and pulleys are massless and also frictionless.

⏭ Relation of acceleration:

✏ Let acceleration of mass m1, m2 and m3 are respectively a1, a2and a3.

$\bigstar \: \underline{ \boxed{ \large\bold{ \sf{ \pink{a_1 + 2a_3 = 2a_2}}}}} \: \bigstar$

⏭ Calculation:

$\bigstar \sf \: \underline{\red{For \: mass \: m_1 \: : }} \\ \\ \mapsto \sf \: T - g = a_1 \: ..... \: (1) \\ \\ \bigstar \: \underline{ \blue{ \sf \: For \: mass \: m_2 \: : }} \\ \\ \mapsto \sf \: 4g - 2T = 4a_2 \\ \\ \mapsto \sf \: 2g - T = 2a_2 \: ..... \: (2) \\ \\ \bigstar \: \underline{ \green{ \sf{For \: mass \: m_3 \: : }}} \\ \\ \mapsto \sf \: 2T - g = a_3 \: ..... \: (3)$ ___________________________________ $\\ \implies \sf \: a_1 + 2a_3 = 2a_2 \\ \\ \implies \sf \: (T - g) + (4T - 2g) = 2g - T \\ \\ \implies \sf \: 6T = 5g \\ \\ \implies \sf \: \boxed{ \purple{ \sf{T = \frac{5g}{6} }}}$ ___________________________________$\\ \leadsto \sf \: putting \: value \: of \: \red{T} \: in \: all \: three \: equations \\ \\ \sf \: we \: get, \\ \\ \bigstar \sf \: \red{a_1 = \frac{g}{6}} \: (upward) \\ \\ \bigstar \sf \: \green{a_2 = \frac{7g}{12}} \: (downward) \\ \\ \bigstar \sf \: \orange{a_3 = \frac{2g}{3} } \: (upward)$

Answer 2

Given :

Let m1 = 1 kg

m2 = 2 kg and

m3 = 3 kg

Suppose the block m1 moves upward with acceleration a1 and the blocks m2 and m3 have relative acceleration a2 due to the difference of weight between them.

Actual acceleration of the blocks m1, m2 and m3 will be

a1, (a1 − a2) and (a1 + a2)

From figure 2, T − 1g − 1a1 = 0    …(i)

From figure 3,

T/2-2g-2(a1-a2)=0    …ii

From figure 4,

T/2-3g-3(a1+a2)=0    …iii

From equations (i) and (ii), eliminating T, we get:

1g + 1a2 = 4g + 4 (a1 + a2)

5a2 − 4a1 = 3g    …(iv)

From equations (ii) and (iii), we get:

2g + 2(a1 − a2) = 3g − 3 (a1 − a2)

5a1 + a2 = g    …(v)

Solving equations (iv) and (v), we get:

a1=2g/29

a2=g-5a1

a2=g-10g/29

=19g/29

Then  a1-a2=2g/29-19g/29

= -17g/ 29

and a1+a2=2g/29+19g/29

=21g/29

So, accelerations of m1, m2 and m3 are

19g/29up, 17g/29 down and 21g/29down, respectively.

Now, u = 0, s = 20 cm = 0.2 m

a2=19g/29        

∴ s=ut+1/2at

⇒0.2=1/2×19/29gt²

⇒t=0.25 s

tried my best

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