A boy of 50 kg is standing in a lift moving down with an acceleration $9.8 m / s^{2}$ . The apparent weight of the boy is:

\frac{50}{9.8} N

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50 \times 9.8 N

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50 N

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Zero

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Solution

The correct option is D Zero
$∵$ The frame of reference of lift is non-internal the boy will feel a pseudo force
$\Rightarrow$ Normal force $+ F_{p s e u d o} =$ Weight
$\Rightarrow$ Apparent weight= Normal force = weight $- F_{p s e u d o}$
$= m g - m a$
$= m \times 9.8 m s^{- 1} - m \times 9.8 m s^{- 1}$
$∴$ Apparent weight $= 0$

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A boy of 50 kg is standing in a lift moving down with an acceleration 9.8m/s2. The apparent weight of the boy is:

The correct option is D Zero∵ The frame of reference of lift is non-internal the boy will feel a pseudo force ⇒ Normal force +Fpseudo= Weight⇒ Apparent weight= Normal force = weight −Fpseudo=mg−ma=m×9.8 ms−1−m×9.8 ms−1 ∴ Apparent weight =0

A boy of 50 kg is standing in a lift moving down with an acceleration $9.8 m / s^{2}$ . The apparent weight of the boy is: