find the acceleration of the 6 kg block in the figure all the surface and Pulley are smooth also strings are inextensible and light. Take g=10 m/s^2
Answers
EXPLANATION.
Acceleration of the 6kg block.
As we know that,
First we draw F.B.D of all the blocks, we get.
Acceleration move from 2kg vertical to 6 kg vertical.
From 6kg block.
⇒ mg - T = ma
⇒ 6g - T = 6a.
⇒ 60 - T = 6a. - - - - - (1).
From 30° splits in the component, we get.
For 2kg inclined block, we get equation.
T + mgsin30° - T₁ = ma.
T + 2 x 10 x 1/2 - T₁ = 2a.
T + 10 - T₁ = 2a. - - - - - (2).
From 2kg block.
⇒ T₁ - mg = ma.
⇒ T₁ - 20 = 2a. - - - - - (3).
⇒ T₁ = 2a + 20. - - - - - (3).
Put the value of equation (3) in equation (2), we get.
⇒ T + 10 - (2a + 20) = 2a.
⇒ T + 10 - 2a - 20 = 2a.
⇒ T - 10 - 2a = 2a.
⇒ T - 10 = 4a. - - - - - (4).
From equation (1) & (4), we get.
⇒ 60 - T = 6a. - - - - - (1).
⇒ T - 10 = 4a. - - - - - (4).
Adding both the equation, we get.
⇒ 60 - 10 = 10a.
⇒ 50 = 10a.
⇒ a = 5m/s².
Acceleration on 6kg block is = 5m/s².
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