A 60, kg man stands on a spring scale in a li | vedclass

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(C) The apparent weight $W'$ of a person in a lift is given by $W' = m(g + a)$,where $a$ is the acceleration of the lift.If the lift is moving upwards

Vedclass · Accepted Answer

The lift,while in constant motion upwards,is stopped suddenly.

Vedclass · Answer

(C) The apparent weight $W'$ of a person in a lift is given by $W' = m(g + a)$,where $a$ is the acceleration of the lift.If the lift is moving upwards with a constant velocity,the acceleration $a = 0$,so the reading is $m 	imes g = 60\, kg$.When the lift suddenly stops while moving upwards,it experiences a downward acceleration (retardation),so $a$ becomes negative. The apparent weight becomes $W' = m(g - a)$.Since the reading decreases from $60\, kg$ to $50\, kg$,it indicates that the lift experienced a downward acceleration (deceleration) while moving upwards.Therefore,we conclude that the lift was moving upwards with constant speed and was suddenly stopped.

$A$ $60\, kg$ man stands on a spring scale in a lift. At some instant,he finds the scale reading has changed from $60\, kg$ to $50\, kg$ for a while and then comes back to the original mark. What should we conclude?

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