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A spring balance is attached to the ceiling of a lift . A man hanga his bag on the spring and the spring reads 49 N when the lift is stationery . If the lift moves downward with an accelration of 5m/s the reading of the spring balance will be

A spring balance is attached to the ceiling of a lift . A man hanga his bag on the spring and the spring reads 49 N when the lift is stationery . If the lift moves downward with an accelration of 5m/s the reading of the spring balance will be

Grade:11

2 Answers

Riddhish Bhalodia
askIITians Faculty 434 Points
8 years ago
due to pseudo force if seen in the frame of reference of the moving elevator the total acceleration on the bag is 9.8 + 5 ~ 15m/s^2 wheres initially it was just 9.8 due to g.
So the weigth now becomes
w = 49*15/9.8 = 75N
arun
123 Points
8 years ago
given that spring balance reads 49N when lift is stationary i.e. . mg=49 . m=49/9.8 . m=5 kg . now lift is accelerating downward so the pseudo force will be in upper direction so the resultant force that spring balance will read will be . R= mg - ma . R= 49 - (5)(5) . R= 24N . so spring will read 24 N

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