A lift is accelerated upward. Will the apparent weight of a person inside the lift increase, decrease or remain the same relative to its real weight? If the lift is going with uniform speed, then?

Put yourself on a weighing scales inside a stationery elevator. So your velocity here is zero meters per second. Let us examine what are the forces acting on your body. We know one of the force is “mg” which acts in the downward direction.The weighing scale will push you up with a certain force which is nothing but the Normal reaction and shows up as weight “W” on the weighing scale (We will take any vector pointing in the upward direction as positive and any vector acting in the downward direction as negative).

We write the forces acting on your body in accordance with Newton's second law of motion which says that the net force acting on you or on a body is equal to its mass into the acceleration induced or

F net = m X a

There are 2 forces acting on the body  (1) mg acting in downward direction and we take it as a negative vector and (2) The normal reaction from the floor lift that is the weight or W (we take it as a positive vector since it points upwards)

So the net force is = W – mg and this should equal the product of resultant acceleration with mass. So the equation that will form is

          F net = W - mg = ma

Or                  W = mg + ma

So if the lift is stationery, a = 0 and W = mg the way it should be. You simply see your weight in the scale

But if the lift starts accelerating up so that velocity increases resulting in positive acceleration “a”. We see that the entity “ma” in above equation adds to mg and therefore the weight W increase.

But if the lift is moving up with a uniform velocity, acceleration “a” will become zero (since acceleration happens only when velocity changes). In such a case again, W = mg or no change

Now imagine that you are reaching the top floor. The lift will start slowing down and this can happen only if a force acts on the lift in the opposite direction. Hence the acceleration “a” now is in downward direction hence negative. So the above equation will now become

W = mg + m X (-a) or

W = mg – ma

So you see when the lift is slowing down at the top, your weight will now reduce by “ma”.