A horizontal force of 129.4 N is applied on a 10 Kg block which rests on a horizontal surface. If the coefficient of friction is 0.3, the acceleration should be

9.8 m / s^{2}

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10 m / s^{2}

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12.6 m / s^{2}

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19.6 m / s^{2}

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Solution

The correct option is B $10 m / s^{2}$
Given,

Mass of block, $M = 10 k g$
coefficient of friction, $μ_{k} = 0.3$

Force applied, $F = 129.4 N$

$F - f_{k} = M a$

$F - μ_{k} M g = M a$

$129 \cdot 4 - M a = μ_{k} m g$

$129 \cdot 4 - 10 \times a = 0.3 \times 10 \times 9 \cdot 8$

$a = \frac{129.4 - 29.4}{10} = 10 m s^{- 2}$ .

Hence, acceleration of block is $10 m s^{- 2}$

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