- #1
iJamJL
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Homework Statement
ASSUME: For all parts of this question, friction is negligible.
A helicopter is used to lift an astronaut of mass m = 69.2 kg a distance s = 11 m vertically out of the ocean by means of a cable. The astronaut rises with constant upward acceleration of magnitude a = 7.44 m/s2 until she reaches the helicopter.
Find Wg, the work done by gravity on the astronaut.
Homework Equations
W=F*distance
F=mg
F=ma
The Attempt at a Solution
I've come to a couple of conclusions, but I think I'm missing a part or two. Here is what I've done.
Wg=F*distance=mg*s
Wg=69.2*9.81*11
That was wrong.
Then I tried this:
If we make gravity the positive direction, then:
F+mg= (-ma)
F= (-mg-ma) = -m(g+a)
Wg= -m(g+a) * s = 13130.7
That was wrong as well, so I'm trying to figure what I'm doing wrong. I have a feeling the answer is somewhere between these two.
Also, I know someone is going to say to draw this out. I always do, but I guess I just think differently or improperly for some reason. I have it drawn like this:
Helicopter is on top, and a person is hanging onto a cable that is 11m long. The person has a mass of 69.2kg, and the acceleration is in the positive y-direction of 7.44 m/s/s. That means that gravity acts in the negative direction, so I guess you could say W= (-mg)*h, but the numerical answer besides the negative sign is still the same.