Question

the lift å moving
2(a) A person of mass 'm' is standing in a
lift
obtain an expression for apparent weight when
(d) Upward with uniform acceleration 'a'
(ii
) downward with uniform acceleration's (azg)
(ii) falling freely and
(ing upward with uniform velocity​

Answer 1

Answer:

Explansoryation:

Answer 2

2(a) Let the weight of the person be ‘m’

When acceleration is upwards, the Normal reaction force is greater than force of mg

According to the second law of motion, net force = Ma

First equation: N - mg = ma

Apparent weight: N = m(a+g)


2) when the lift is moving down, mg > N

So according to second law of motion,


Mg - N = ma
M(g-a) = N

When body is freely falling, it is in a state of weightlessness and has no apparent weight as

N = m(g-g)

4) when body is moving the uniform velocity, there is no accelerated motion
Hence

N = mg