Question

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

• A. \(\frac{4g}{15}\)
• B. \(\frac{2g}{15}\)
• C. \(\frac{g}{15}\)
• D. \(\frac{2g}{3}\)

Answer 1

The Correct Option is A

Acceleration of the 2 kg block: The 2 kg block, as depicted in the diagram, has an acceleration described by the equation \(T - 2g \sin 37^\circ = 2a\) (i). For the 4 kg block, the equation is \(4g - 2T = 4a\), which simplifies to \(g - T = a\) (ii).

\(2g - a - 2g \times \frac{3}{5} = 2a\)

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

The correct option is (A): \(\frac{4g}{15}\).