A horizontal force of $1 g w t$ . is applied on a mass of $9.8 k g$ placed on a smooth surface. The acceleration produced is:

1 m s^{- 2}

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\frac{1}{9.8} m s^{- 2}

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9.8 m s^{- 2}

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10^{- 3} m s^{- 2}

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Solution

The correct option is D $10^{- 3} m s^{- 2}$
Given,
$F = 1 g w t$
$m = 9.8 k g$
Force, $F = m a$
$1 g w t = 9.8 a$
$\frac{1}{1000} \times 9.8 = 9.8 a$
$a = \frac{1}{1000} = 10^{- 3} m / s^{2}$
The correct option is D.

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