Two blocks with masses M1 and M2 of 10kg and 20kg respectively are placed as in figures- Coefficient of friction -0-2 between all surfaces- then tension in string and acceleration of M2 block will be

Let us consider from the figure #160;Due to block M_1 the force acting on vertical direction is T sin 30^∘=10 Kg #160; #160;—– #160; ( ...

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Standard XII

Physics

Constrained Motion : General Approach

Two blocks wi...

Question

Two blocks with masses $M_{1}$ and $M_{2}$ of $10 k g$ and $20 k g$ respectively are placed as in figures. Coefficient of friction $μ = 0.2$ between all surfaces, then tension in string and acceleration of $M_{2}$ block will be

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Solution

Let us consider from the figure

Due to block $M_{1}$ the force acting on vertical direction is $T sin 30^{\circ} = 10 K g$ ----- $(i)$

therefore, $T = 196 N$

For block $M_{1}$ horizontal component of tension

force of normal reaction $N = T cos 30^{\circ} = 196 \times \frac{\sqrt{3}}{2} = 170$

For block $M_{2}$ , vertical direction force are written as

$20 g - 2 \times μ N = 20 \times a$

now substitute the values of $μ = 0.2, N = 170$

$(20 \times 9.8) - (2 \times 0.2 \times 170) = 20 \times a$

$20 a = 196 - 68$

$20 a = 128$

$a = \frac{128}{20} = 6.4 m / s^{2}$

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Two blocks with masses M1 and M2 of 10kg and 20kg respectively are placed as in figures. Coefficient of friction μ=0.2 between all surfaces, then tension in string and acceleration of M2 block will be

Two blocks with masses $M_{1}$ and $M_{2}$ of $10 k g$ and $20 k g$ respectively are placed as in figures. Coefficient of friction $μ = 0.2$ between all surfaces, then tension in string and acceleration of $M_{2}$ block will be