Question:

Find the acceleration of \(2\) \(kg\) block shown in the diagram (neglect friction)
Find the acceleration of 2kg  block

Updated On: Nov 7, 2024
  • \(\frac{4g}{15}\)
  • \(\frac{2g}{15}\)
  • \(\frac{g}{15}\)
  • \(\frac{2g}{3}\)
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The Correct Option is A

Solution and Explanation

The acceleration of 2 kg block is : 
The 2 kg block as shown in the diagram 
\(T-2g\) ; \(sin\) \(37 = 2a\)…(i)
For the \(4\) kg block 
\(4g-2T\)
\(g-T = a\)….(ii)

\(2g-a-2g\) \(\times \frac{3}{5}=2a\)

\(a = \frac{4g}{15}\)

The Correct Option is (A): \(\frac{4g}{15}\)

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Concepts Used:

Acceleration

In the real world, everything is always in motion. Objects move at a variable or a constant speed. When someone steps on the accelerator or applies brakes on a car, the speed of the car increases or decreases and the direction of the car changes. In physics, these changes in velocity or directional magnitude of a moving object are represented by acceleration

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