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Find the Acceleration of the Block of Mass M in the Situation Shown in Figure (5−E15). All the Surfaces Are Frictionless and the Pulleys and the String Are Light. - Physics

Sum

Find the acceleration of the block of mass M in the situation shown in the following figure. All the surfaces are frictionless and the pulleys and the string are light.

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Solution

The free-body diagram of the system is shown below:

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
\[\Rightarrow a = \frac{g}{6}\]
Hence, the acceleration of mass
\[M = 2a = 2 \times \frac{g}{6} = \frac{g}{3} \left (\text{ up the plane }\right) .\]

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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 5 Newton's Laws of Motion
Q 32 | Page 81
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