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ICSE Class 10 Answered

  2)A person is standing in an elevator.In which case he finds his weight less than actual a)the lift moves up with constant acceleration  b)the lift moves down with constant  acceleration  c)the lift moves up with uniform velocity  d)the lift moves down with constant velocity  Explain each one
Asked by Sunita | 28 Feb, 2018, 12:25: PM
answered-by-expert Expert Answer
There are two forces (1) Force exerted on passenger by Scale (upward direction) and (2) passenger's weight due to gravitation
 
as per newton's second law   Fps-W = m×a; where a is the net acceleration
 
Fps = W+ma = mg+ma = m(g+a)
 
Weight measuring machine gives the reading due to Fps.
 
(a) when lift moves up with constant acceleration
 
Fps = m(g+a) ; hence passenger's weight is more than actual as per weighing machine reading.
 
(b) when lift moves down with constant acceleration
 
Fps = m(g-a) ; passenger's weight is less than actual as per weighing machine reading
 
(c) & (d) when lft moves with constant velocity acceleration is zero. 
hence Fps = mg ; true weight
Answered by Thiyagarajan K | 28 Feb, 2018, 03:16: PM

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