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A spring weghing machine inside a statio...

A spring weghing machine inside a stationary life reads 50 kg when man standds on it. What would happen to the scale reasing if the life is moving upward with (i) costant velocity, and (ii) constant acceleraton ?

Text Solution

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(i) In the case of constant velocity of lifr, there is no fictitious force, therefore the apparent weight = actual weight. Hence the reading of machine is 50 kgwt.
(ii) In this case the acceleration is upward, the fictitious force ma acts downward, therefore apparent weight is more than actual weight i.e. W' = m (g + a). Hence acale shows a reading = m (g + a) `=mg(1+(a)/(g))/(g)=(50+(50a)/(g))kg wt.`
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