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A body of mass 40 kg stands on a weighing machine in an accelerated lift. The reading on the scale of the weighing machine is 300 N. Find the magnitude and direction of acceleration. $(g = 9 .8 {ms}^{- 2})$

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2.3 {m/s}^{2} downward

4.6 {m/s}^{2} upward

4.6 {m/s}^{2} down ward

2.3 {m/s}^{2} upward

answer is B.

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Detailed Solution

m = 40 kg, w = 300 N

$w = m (g - a)$ (lift moving down ward)

$300 = 40 (g - a)$

$\frac{300}{40} = g - a$

$a = 9 .8 - \frac{30}{4}$

$= \frac{39.2 - 30}{4}$

$= \frac{9.2}{4}$

$= 2.3 {m/s}^{2} downward$

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