'A 60.kg person is standing in a lift. What would be their apparent weight as the lift is moving downwards and decelerating to rest at 2.40ms^2?'. I have already answered this question however I am not sure that it is correct.
My Answer:
Let w = weight due to gravity
Let n = apparent weight
ΣF=ma
Therefore n-w = ma
n = ma +w
Therefore n = ma + mg
= m(a + g)
= 60 (-2.4 + 9.8)
= 444N
= 45.27 kg
Is this right or should it end up being n = m(a - g) Remembering it is decelerating downwards to a rest
Thanks

Question

'A 60.kg person is standing in a lift. What would be their apparent weight as the lift is moving downwards and decelerating to rest at 2.40ms^2?'. I have already answered this question however I am not sure that it is correct. 
My Answer:
Let w = weight due to gravity
Let n = apparent weight 
ΣF=ma
Therefore n-w = ma
n = ma +w 
Therefore n = ma + mg
= m(a + g)
= 60 (-2.4 + 9.8)
= 444N
= 45.27 kg
Is this right or should it end up being n = m(a - g) Remembering it is decelerating downwards to a rest
Thanks

hugsnotughs · Answer

I'm not sure.  With the equation solving you have it looks right, though.  But I can't tell you fi you're doing it wrong or right.  Best of luck, though! (^_^)

anonymous · Answer

Thank you, I am just not sure about the + or - does anyone know if in an elevator decelerating downwards you feel lighter or heavier? I can work it out from that.

hugsnotughs · Answer

@mashy can help, but not sure if he's on.  That's the only person I know who's good in Physics for now.  And it's late, at least in some of the US, and most people aren't on at these times.  But I hope someone comes soon. :)

anonymous · Answer

Thanks

vincent-lyon.fr · Answer

This is not correct.

Remember  $\vec W+\vec N = m\vec a$
$\vec W$ is down and $\vec N$ is up.

Now the only question is whether $\vec a$ is up or down.
What do you think?

anonymous · Answer

yea like @Vincent-Lyon.Fr  mentioned.. u got the direction of the acceleration wrong ;-).. 

@hugsnotughs.. thanks hon :D