the apparent weight of man inside a lift moving up with certain acceleration is 900N.when the lift is coming down with the same acceleration apparent weight is found to be 300N.the mass of the man is
Answers
Weight while coming down =300N
So the weight of the men = 600N
Now w=mg
m=w/g
= 600/10 (taking. g = 10m/s square)
= 60kg
Therefore the mass of the man is 60 kg.
Given:
The apparent weight of the man inside a lift moving up with certain acceleration = 900 N
The apparent weight of the man inside a lift coming down with the same acceleration = 300 N
To Find:
Mass of the man
Solution:
This question can be simply solved using the below-shown approach.
Apparent weight is nothing but the Normal reaction of the man inside the lift.
⇒ Apparent Weight = Normal Reaction
Case-1: Lift moving up:
Normal reaction N = mg + ma = 900_Eq-I ( Given )
Where m = mass of the man
g = acceleration due to gravity = 10 m/s²
a = acceleration of the man
Case-2: Lift moving down:
Normal reaction N = mg - ma = 300_Eq-II ( Given )
Solving equations I and II, we get,
mg + ma = 900
mg - ma = 300
______________
2mg + 0 = 1200
⇒ mg = 600 ⇒ m = 600/10 = 60 kg
Therefore the mass of the man is 60 kg.
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