Physics, asked by Anonymous, 4 months ago

an object of mass M is kept on the floor of a lift what will be the Apparent weight of the object when the lift moves from a constant speed V downwards

Answers

Answered by dualadmire
2

Given:

The mass of the object is = M

The lift is moving with a constant velocity in downwards direction = V

To find:

The apparent weight of the object when it is kept on the floor of the lift.

Solution:

The apparent weight can be found out by vector addition of the weight of the object and the other forces acting on the object.

Since the lift moves with a constant velocity, its acceleration = 0

Apparent weight = Mg

If the lift had an acceleration a in downward direction then,

Apparent weight must had been = M(g - a)

Therefore Apparent weight of the object = Mg.

Answered by nirman95
4

Given:

An object of mass M is kept on the floor of a lift.

To find:

Apparent weight of the object when lift is moving down with constant velocity v ?

Calculation:

First of all , what is APPARENT WEIGHT inside the lift?

  • Apparent weight is the the new normal reaction experienced by the person (in the lift), when the lift is accelerated downwards/upwards.

  • For example: when lift is being accelerated downwards, the man will experience a pseudo-force upwards, and his normal reaction will be reduced than his weight. So, he will feel somewhat less weight.

  • Refer to diagram.

 \therefore \: F_{app} = mg - ma

 \implies\: F_{app} = m(g - a)

Now, since question says that lift is moving with constant velocity v , the value of "a" = 0 m/s².

 \implies\: F_{app} = m(g - 0)

 \implies\: F_{app} = mg

 \implies\: F_{app} =weight \: of \: body

So, apparent weight will be equal to weight of the body.

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