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Standard XII

Physics

Relative Position

A 2kg block...

Question

A $2 k g$ block is placed over a $4 k g$ block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is $0.20$ . The acceleration of the two blocks if a horizontal force of $12 N$ is applied to the upper block $(g = 10 m s^{- 2}$ )

4 m s^{- 2}, 2 m s^{- 2}

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2 m s^{- 2}, 2 m s^{- 2}

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3 m s^{- 2}, 3 m s^{- 2}

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4 m s^{- 2}, 1 m s^{- 2}

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Solution

The correct option is C $4 m s^{- 2}, 1 m s^{- 2}$
The correct option is D.

Given,

$m a s s e s = 2 k g, 4 k g$

Coefficient $0.20$

Force $= 12 N$

$R_{1} - 2 g = 0$

$R_{1} = 2 x 10 = 20$

$2 a + 0.2 R_{1} - 12 = 0$

$2 a + 0.2 (20) = 12$

$2 a = 12 - 4$

$a = 4 m / . s^{2}$

Again,

$4 a - μ R_{1} = 0.2 (20) = 4$

$a_{1} = 1 m / s^{2}$

$2 k g$ block has acceleration $4 m / s^{2}$ and that of $4 k g$ is $1 m / s^{2}$

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A 2 kg block is placed over a 4 kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.20. Find the acceleration of the two blocks if a horizontal force of 12 N is applied to the upper block. Take g = 10 $m / s^{2}$ .

Q. A 2 kg block is placed over a 4 kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the block is 0.20. Find the acceleration of the two blocks if a horizontal force of 12 N is applied to (a) the upper block, (b) the lower block. Take g = 10 m/s².

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A 2kg block is placed over a 4kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.20. The acceleration of the two blocks if a horizontal force of 12N is applied to the upper block (g=10ms−2)

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A $2 k g$ block is placed over a $4 k g$ block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is $0.20$ . The acceleration of the two blocks if a horizontal force of $12 N$ is applied to the upper block $(g = 10 m s^{- 2}$ )