# Find the acceleration of the block of mass M in the situation shown in figure (5-E15). Ail the surfaces are frictionless and the pulleys and the string are light.

Given: We can calculate by the FBD

Let acceleration of the block of mass 2M be a.

So, acceleration of the block of mass M will be 2a.

M(2a) + Mgsinθ − T = 0

T = 2Ma + Mgsinθ (i)

2T + 2Ma − 2Mg = 0

From equation (i), 2(2Ma + Mgsinθ) + 2Ma − 2Mg = 0

4Ma + 2Mgsinθ + 2Ma − Mg = 0

6Ma + 2Mgsin30° + 2Mg = 0

6Ma = Mg

Hence, the acceleration of mass

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