# EK 30 min exam-weight & acceleration

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#### sacha

##### Full Member
10+ Year Member
In the EK 30 min exam for lecture 3, #64, the answer key states the apparent weight of a pilot increases while pulling out of a dive since he is decelerating.

In the answer key: 'the apparent weight of the pilot is the reading on a scale if he were sitting on a scale during his dive. this means that the scale underneath him would have to push up with more force than his weight.' I don't understand why this last statement is true. Anyone?

i think this problem is analgous to a person being accelerated in an elevator so it might help if you think about it like that.

it says that the pilot is pulling out of a dive, meaning that he is no longer "free-falling" but pointing (or trying to) the nose of the plane against gravity. you can imagine this in an evelator that is constantly accerlating you against gravity.

if you do a free body dia. and do fnet=ma you'll see that the normal force (thus the apparent weight) will increase when going against gravity. so if you assume this evelator is going up 3 m/s^2 then you'll have the equation: N-mg=ma, where N is the normal force that youre trying to solve for. you get N= m(10+3) so the normal force is higher than if just free-falling or "diving".

in the same case if the pilot decided to accelerate while diving, or the elevator was accelerated downward with a force outside gravity, the apparent weight would decrease. N-mg=m(-a) -> N= m(g-a) so you would weigh less if you were in this elevator. hope that helps!

i think this problem is analgous to a person being accelerated in an elevator so it might help if you think about it like that.

it says that the pilot is pulling out of a dive, meaning that he is no longer "free-falling" but pointing (or trying to) the nose of the plane against gravity. you can imagine this in an evelator that is constantly accerlating you against gravity.

if you do a free body dia. and do fnet=ma you'll see that the normal force (thus the apparent weight) will increase when going against gravity. so if you assume this evelator is going up 3 m/s^2 then you'll have the equation: N-mg=ma, where N is the normal force that youre trying to solve for. you get N= m(10+3) so the normal force is higher than if just free-falling or "diving".

in the same case if the pilot decided to accelerate while diving, or the elevator was accelerated downward with a force outside gravity, the apparent weight would decrease. N-mg=m(-a) -> N= m(g-a) so you would weigh less if you were in this elevator. hope that helps!

understood! thanks alot