Find the acceleration of \(2\) \(kg\) block shown in the diagram (neglect friction)
- \(\frac{4g}{15}\)
- \(\frac{2g}{15}\)
- \(\frac{g}{15}\)
- \(\frac{2g}{3}\)
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:
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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.
