physics question

[Gibson486]

I nee dhelp w/ these physics questions for my hw.

1a) What is the acceleration of two falling sky divers (mass 105. kg including parachute) when the upward force of air resistance is equal to 1/4 their weight?

A:I got 7.25 m/s^2 for this one

1b) After popping open the parachute, th edivers descend lesuirely to the ground at contsant speed. What now is the force of air resisitence on te sky divers and their pararchutes?

A; This is the one I am having trouble with. i think that it would be the differenc eof the force from the net force of teh divers and the ari resisitance?

2)An exceptional standing jump would raise a person .68m off the ground. To do this, what force must a 69 kg peron exert against the ground? Assume that the person crouches a distance of .22 m prior to jumping, and thus the upward force has this distance to act over before he leaves the ground.

A: This is the one I am really having trouble with. I have no idea about this one.

Don't just give me teh answers. Explain them please, especially, the last one.

 

[silverpig]

1A - 7.35 m/s^2

basically you have a force of Fg down, and Fg/4 up, for a total of .75 Fg down. Basically you just take .75 x g... I get 7.35 m/s^2 if g = 9.8.

1B

Since they are at constant velocity, F(net) = 0. So air resistance must equal their weight = g x m = 9.8 x 105 = 1029 N

2A

Will post answer in a minute. Gonna take a bit of number crunching.

 

[Tdawg951]

haha, the second one might use energy, since its force over a distance

 

[silverpig]

2A is a multistep problem. Break it up into jumping force and flight.

Figure out what the person's velocity must be for him to go up to .68 m and back down. Once you know that initial velocity requirement, you can use that as your final velocity for the first part of the problem...

sorry I can't finish this right now, I'm sure someone else can... I'm walking someone through a linux install right now

 

[silverpig]

Originally posted by: Tdawg951
haha, the second one might use energy, since its force over a distance

Actually energy would be an easier way to do it rather than kinematics.

Find the potential energy at the top of the jump h = .68... = mg = 69 x 9.8 = 676.2 J
At h = 0, potential energy = 0, and kinetic energy = 676.2 J

so now you have 1/2 m v^2 = 676.2 J

You know m, so figure out v.

This'll be your final velocity for the jumping stage.

 

[RaynorWolfcastle]

here's the gist of how to solve 2A

Consider that once the person is off the ground, the only force being applied is gravity.
Because of this, you know that the person has to have an initial velocity such that he can reach .68m despite gravity.

so you use v[f]^2 = v[o]^2 +2a*dist
we know that v[f] = 0,
a = - 10 m/s^2,
dist = .68m

Now, solve the equation for v[o]

so you know how fast he has to be going to jump this high now.

again use v'[f]^2 = v'[o]^2 +2a*dist
but we know that
v'[f] = v[o] <-- watch those primes
v'[o] = 0
dist = 0.22m

now solve for a

now do F=ma and boom, you've got the force (assuming of course it's constant)

 

[RaynorWolfcastle]

Originally posted by: Tdawg951
haha, the second one might use energy, since its force over a distance

yeah, you could definetly use work and energy but you don't have to (directly anyways)... see my solution

 

[silverpig]

for jumping stage:

Vf^2 = Vo^2 + 2ad

since f = ma

Vf^2 = Vo^2 + 2fd/m

vf = what you just figured out
vo = 0
f = what you want to find
d = .22
m = 69 kg

 

[Rahminator]

I did #2. It's a simple kinematics problem. Basically, you have to figure out average acceleration during the .22m it takes to jump up off the ground. You know that at the highest point, final velocity is 0 and that total flight is .68m. To figure out initial speed at which you leave the ground:

Vf^2 = Vi^2 - 2gx

Solving for Vi you get 3.65 m/s. So that is the speed at which you leave the ground. However, the person crouched for .22m before he left the ground at 3.65 m/s. What you have to do is figure out the acceleration of the person during the .22m distance. The initial speed is 0 (crouching) and the final speed is 3.65 m/s (leaving the ground) so you use the same equation as above, but this time solving for a:

a = (Vf^2 - Vi^2) / (2x)

and you get 30.27 m/s/s. F = ma so F = 2089.2 N.

 

[Gibson486]

I got 572.7 N. I am going to check to see if it is right.

 

[Gibson486]

the answer to #2 is not 572 or 2089!

 

[RaynorWolfcastle]

Originally posted by: Gibson486
the answer to #2 is not 572 or 2089!

hmmm... we calculated Fnet, maybe they wanted Fapplied which would be Fnet - weight where weight should be in the opposite direction (remember that an applied force equal and opposite to his weight gets him no velocity, just weightlessness (sp?) )

Fapplied which works out to be ~2769 N

also, make sure that you use the right value for g... I used g = 9.81, they may expect you to use g = 10

 

[Gibson486]

Yup, they wanted Force applied I did teh calculations and got 2765 and it was right Now, the only problem is this one I thought it was right, but it's wrong:

A person stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.63 of the person's regular weight. Calculate the acceleration of the elevator, and find the direction of acceleration.

I keep getting 6.17 m/s/s and another time i got 2.65 m/s/s. Both were wrong. the 2.65 m/s/s I got in class while working in a group. The 6.17 I got while working by my self. I did teh calculations it looks like that 6.17 does give 63% of the person weignts. I am stumped on this. To get these solutions, all I did was take 63% of the persons weight and just substituted it into the F=ma equation, with mass staying the same ( I ofcrouse picked my own weight since weight was not given).


BTW, to prevent plagerism, these questions come from "Gianlco (sp?) physics for engineers and scientists".

 

[silverpig]

The answer for that one is 9.81 - 6.17 = 3.64 m/s^2 downwards.

If he feels only 0.63 of his normal weight, then only 0.63 of g is pushing him in contact with the floor of the elevator. That value is 6.17. The "missing" part of g is due to the elevator accelerating downwards. This comes out to be 3.64 m/s^2


Sorry about the horrible explanation, but it's been a while since I've had to actually explain this stuff.

 

[Gibson486]

Thanks, that is what I thought it was, but since it was 63% of twh eight, I thought it was acclerating 63% of gravity. I just got confused. I wa sgonna go to the tutors in my school to help me but it sucks b/c evertime I go, there are always 10 peopel in there sop I end up learning nothing. Thanks You guys think it will be alright if i post any other question sin the future i have trouble with?