Question

A spring weighing machine inside a stationary lift reads 50kg when a man stands on it. What would happen to the scale reading if the lift is moving upward with 1) constant velocity, and 2) constant acceleration??

Answer 1

Hey...mate here is yur ans...
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1) In the case of constant velocity of lift, there is no fictitious force; therefore the apparent weight = actual wieght. Hence the reading of machine is 50 kgwt.

2) In this case the acceleration is upward, the fictitious force R = ma acts downward, therefore apparent weight is more than actual weight that is W' = W + R = m (g+a).

Hence scale shows a reading = m (g+a) = mg [1+a/g] N=[50+50a/g] kgwt.

Hope it helps u✌

Answer 2

in the case of constant velocity of lift there is no fictitious force; therefore the apparent weight = actual weight.
Hence the reading of machine is 50 kgwt.

Hope it will help you