A person of mass 'm' is standing in a lift. Find his apparent weight when lift is (show mathematical
calculation)
1.moving upward with uniform acceleration 'a'.
2.moveing downward with uniform acceleration 'a' (a<g)
3.falls freely
Answers
Answer:
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Explanation:
1. mg+ma
2. mg-ma
3. When lift falls freely it's apperent weight is zero.
Because of its free fall.
The apparent weight of the person-
1. when the lift is moving upward acceleration- m(g+a)
2. moving downward with uniform acceleration m(g-a)
3. falls freely - zero.
GIVEN
Mass of a person = m
1. lift moving upward with acceleration a
2. lift moving downward with acceleration a.
TO FIND
Apparent Weight of the person.
SOLUTION
(1) LIFT MOVING UPWARD WITH UNIFORM ACCELERATION :
when the lift is in rest position,
The force by the person on the lift is in the downward direction.
The force on the lift by the person = Weight of person
when the lift is at rest,
weight of person = m×g (g=acceleration due to gravity
When Lift Is Moving In Upward Direction-
when lift starts moving in an upward direction, a Reaction force is applied to the person opposite to the direction of acceleration.
In this case, the Direction of Reaction force is in the downward direction.
Reaction force, F = m × a
Apparent weight = Initial weight + Reaction force
= mg + ma
Apparent weight = m(g+a)
Hence, The apparent weight of the person when the lift is moving in the upward direction is m(g+a).
(2) LIFT MOVING DOWNWARD WITH UNIFORM ACCELERATION :
when the lift starts moving in a downward direction, a Reaction force is applied to the person opposite to the direction of acceleration.
In this case, the Direction of reaction force is in an upward direction.
Reaction force, F = m × a
Apparent weight = Initial weight - Reaction force
= mg - ma
Apparent weight = m(g-a)
(3) FALLS FREELY:
When the lift falls freely.
There is no reaction force that the body can exert on the left.
Hence, the apparent weight of a person under free fall is zero.
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