A man in a lift feels an apparent weight W when the lift is moving up with a uniform acceleration of 1/3 rd of the acceleration due to gravity. If the same man were in the same lift now moving down with a uniform acceleration that is 1/2 of the acceleration due to gravity, then what is his apparent weight?
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# Answer- 3W/8
# Explaination-
a) When lift is going upwards-
Acceleration is downwards.
Total acceleration = g + g/3 = 4g/3
Apparent weight = ma = 4mg/3 = W
b) When lift is going downwards-
Acceleration is dupwards.
Total acceleration = g - g/2 = g/2
Apparent weight = ma = mg/2
Apparentl weight = 3/8 (4mg/3) = 3W/8
Hence, apparent weight in lift going down is 3W/8.
Hope that was useful...
# Answer- 3W/8
# Explaination-
a) When lift is going upwards-
Acceleration is downwards.
Total acceleration = g + g/3 = 4g/3
Apparent weight = ma = 4mg/3 = W
b) When lift is going downwards-
Acceleration is dupwards.
Total acceleration = g - g/2 = g/2
Apparent weight = ma = mg/2
Apparentl weight = 3/8 (4mg/3) = 3W/8
Hence, apparent weight in lift going down is 3W/8.
Hope that was useful...
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