Question

find acceleration of block and tension in strings if m1 =8kg m2= 4kg and take g=10m/s²​

Answer 1

EXPLANATION.

=> Mass of block¹ = m¹ = 8kg

=> Mass of block² = m² = 4kg

=> g = 10m/s²

To find the acceleration and tension

in string.

=> As we know that first we draw F.B.D of block

=> then the block¹ = m¹ moves downward and

acceleration is also move downward in the

direction of m¹.

=> equation will be written as.

=> 8g - T = 8a ....... (1)

=> T - 4g = 4a ....... (2)

From equation (1) and (2) we get,

=> 4g = 12a

=> 4 X 10 = 12a

=> 40 = 12a

=> a = 40/12 = 10/3 m/s²

put the value of a = 10/3 in equation (2) we get,

=> T - 4g = 4 X 10/3

=> T - 40 = 40/3

=> T = 40/3 + 40

=> T = 40 + 120 / 3

=> T = 160/3 N

Therefore,

=> Acceleration = 10/3 m/s².

=> Tension = 160/3 N.

Answer 2

Answer:

✳️ Given ✳️

• $\sf{m_1}$ = 8kg
• $\sf{m_2}$ = 4kg
• g = 10m/s²

✳️ To Find ✳️

• Acceleration of block
• Tension in strings

✳️ Solution ✳️

✒️ Let, the two equation will be ,

$\leadsto$ 8g - T = 8a ..... Equation no (1)

$\leadsto$ T - 4g = 4a ..... Equation no (2)

➕ Putting the value of equation no (1) and (2) we get,

$\implies$ 4g = 12a

$\implies$ 4 × 10 = 12a

$\implies$ 40 = 12a

$\implies$ a = $\sf\dfrac{\cancel{40}}{\cancel{12}}$

$\dashrightarrow$ a = $\dfrac{10}{3}$

Again, we have to putting the value of a in the equation no (2) we get,

$\implies$ T - 4g = 4 × $\dfrac{10}{3}$

$\implies$ T - 4 × 10 = 4 × $\dfrac{10}{3}$

$\implies$ T - 40 = $\dfrac{40}{3}$

$\implies$ T = $\dfrac{40}{3}$ + 40

$\implies$ T = $\dfrac{40 + 120}{3}$

$\dashrightarrow$ T = $\dfrac{160}{3}$ N

$\therefore$ Acceleration of block = $\dfrac{10}{3}$ m/s².

And, Tension in strings = $\dfrac{160}{3}$ N

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