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Determine the magnitude of the apparent weight of a 56 kg student standing in an elevator when the elevator is experiencing an acceleration of (a) 3.2 m/s^2 downward and (b) 3.2 m/s^2 upward.

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W = m*a When it's going up, acceleration is: g+a, when it's going down acceleration is g-a
a) total a = -9.8-3.2 = -13 W = 56*(-13)N
You see, here's what I did for a and b.

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Other answers:

b) total a = -9.8 + 3.2 = -6.6 W = 56*(-6.6)N
a) f=ma =(56)(3.2+9.81) =728.56N
Why is the force negative in both cases?
because weight is pushing down.
Ok. Give me a second to redo this...
same thing, just change the signs
you found the normal force, which has the same magnitude, but opposite direction
And its alright if my signs are negative? Because the answers are A) 3.7*10^2N and b) 7.3*10^2N
ohh yea they asked for magnitude. If they just asked for weight, ur answers wld be negative
since weight has magnitude and direction. In this case it's going down, hence the negative.
k. ill work with this. if i need help, ill call out
the answer for A) was wrong. you probably explained this, but i forgot
We got -728.56N. The book says 3.7*10^2N
Wait in the answer, is the answer for a greater than the answer for b?
A) 3.7*10^2N and b) 7.3*10^2N Aren't those the answers?
oh yea.. obv. Hang on I know what we did wrong
for A), i did this: "56)(-3.2-9.81)
well it makes sense.. i just always confuse the signs of g. g has to be positive in this case.
would that make my value of 3.2N positive too?
it's the sign of g, i always get confused with it..
am I to consider 3.2N a positive or negative force?
well is it pushing down or up?
It's pushing down
If it was -ve, it would make sense mathematically, but not in terms of real world physics(for me)
then it's negative, but they're just asking for magnitude, or abs value of that force.
ok. give me a moment to readi guess

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