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A $20 kg$ body is pushed with just enough force to start it moving across a floor and the same force continues to act afterwards. The coefficient of static and kinetic friction are $0.6$ and $0.2$ respectively. The acceleration of the body is :- $(g = 10 m / s^{2})$

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$4 m / s^{2}$

$1 m / s^{2}$

$2 m / s^{2}$

$3 m / s^{2}$

answer is D.

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Detailed Solution

$f_{L} = μ_{s} N = 0.6 \times 20 g = 120 N$

$Now, a = \frac{120 - 0.2 \times 20 g}{20} = 4 m / s^{2}$

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