Two Questions, one very complicated

[KingNothing]

Please help me with these two questions...I kinda know where to start, but If I jsut saw it worked out once I could do all of them of this style, so please help.

Question 1:
"A 65-kg paratrooper is falling, and hits the ground at 4 m/s. He makes an indentation 0.2 m deep in the ground. What force does he hit the ground with?"

Question 2: This one is complicated:
A 4000 kg train car is atop a 5 degree incline which is 5 m high (from baseline to top of hill, making it about 57 m long). It begins rolling down. The coefficient of the rolling frciton is 0.05 which is also static friction. At the bottom of the incline, it attaches to 3 identical carts. What is the final velocity of all 4 carts (when they form a train)?

[ShawnD]

Originally posted by Decker
Question 1:
"A 65-kg paratrooper is falling, and hits the ground at 4 m/s. He makes an indentation 0.2 m deep in the ground. What force does he hit the ground with?"

3 energies are at work. The ground does work on the man over a distance, gravity does work over a distance (gravity is often forgotten), and he had an initial kinetic energy.

F_{ground}d = \frac{1}{2}mv^2 + F_{gravity}d

F_{ground} = \frac{\frac{1}{2}mv^2 + mgd}{d}

Question 2: This one is complicated:
A 4000 kg train car is atop a 5 degree incline which is 5 m high (from baseline to top of hill, making it about 57 m long). It begins rolling down. The coefficient of the rolling frciton is 0.05 which is also static friction. At the bottom of the incline, it attaches to 3 identical carts. What is the final velocity of all 4 carts (when they form a train)?

Write an equation for the energies
potential energy = friction + kinetic energy

mgh = Fd + \frac{1}{2}mv^2

mgh = \mu mg \cos(\theta)d + \frac{1}{2}mv^2

Then isolate for V. It will be easier if you just factor out the mass.


To find the velocity of all the carts, use the formula for momentum

m_1v_1 = m_2v_2

[KingNothing]

It looks good...but we haven't done anything yet with kinetic energy or 'work'. I don't even know what h is in the second one.

[KingNothing]

mgh=Fd+.5mv^2
(9.8)(5)=(0.05)(0.996)(57)+.5(v^2)
49=2.8386+.5v^2
92.3228=v^2
v=9.6 m/s?

Does that sound right?
Then the whole train would be going 2.4 m/s when they connect?

I still need lot of help on the other one, I don't know what work done by the ground is!

[Integral]

Perhaps we are making the dent in the ground problem to hard, the dent is a red herring, the paratrooper is FALLING therefore his acceleration = g so F= mg end of story.

[ShawnD]

Integral, that's not true at all. Ever notice that you can stand straight up but if you jump down from something maybe a meter high, you always need to bend your legs?
Actually try it.

[Chen]

Originally posted by Integral
Perhaps we are making the dent in the ground problem to hard, the dent is a red herring, the paratrooper is FALLING therefore his acceleration = g so F= mg end of story.
I'll try to explain why that's not true... when you hit the ground, you feel the normal force, right? Usually the normal force is equal to your weight so you feel no acceleration:
\Sigma F = mg - N = 0
However, when you hit the ground with speed, the normal force is a lot stronger than your weight. If it were still equal to your weight, your acceleration would be zero and the speed with which you hit the ground will not go away (and of course, after the impact your velocity is zero so this can't be true).

[cookiemonster]

What's the normal force got to do with anything?

cookiemonster

[Doc Al]

What's the normal force got to do with anything?

cookiemonster
Everything! That's what's stopping the paratrooper.

The best one can do with this problem is find the average force exerted by the ground in stopping the paratrooper within a distance d. ShawnD did that several posts ago.

[Chen]

What's the normal force got to do with anything?

cookiemonster
A lot... [:)] Do you have a scale around? Place it on the ground and step up on it. It will show your "real" weight, right? Now instead of stepping up on the scale, try jumping on it (without breaking it of course... don't try this if you ate too many cookies this morning!) while looking at its display. You will see that for a brief moment the scale, which measures the normal force, shows an extremely large number.

[mrbill]

As far as jumping off a chair and having to bend your legs, I believe that would fall under momentum, not force. p=mv so the increased velocity on impact when jumping from any given height would increase, increasing one's momentum. f = p/t so the ground would have to exert a greater force on the person in order to stop. By bending your legs you would be increasing t, reducing the force exerted on the body.

mrbill

[nuub]

can you assume that the deceleration as he hits the ground is constant?

If so use the formula v^2=u^2+2*a*s,
This will let you work out the acceleration and so using F=ma you can work out his force.

therefore 4^2/(2*0.2) =40
f=65*40=2600N

This looks a bit big so I may be wrong

[nuub]

for the other question f=mgsin5-0.05*mgcos5
f=ma therefore you can work out the acceleration of the single truck
then again use v^2=u^2+2as to work out the final speed
then use conservation of momentum at the bottom
i.e. mu=4mv and you can work out the final speed of all the cars

[cookiemonster]

I certainly hope the paratrooper doesn't hit the ground with the same force as his normal force! He'd plow right through the ground! The normal force had better be significantly bigger than the impact force, but that still doesn't answer my question. What does the normal force have to do with the force with which the paratrooper hits the ground? The normal force only has to do with how quickly the paratrooper stops, no?

cookiemonster

[Doc Al]

I certainly hope the paratrooper doesn't hit the ground with the same force as his normal force!
The force that the ground hits the paratrooper is the normal force, by definition. And the paratrooper had better hit the ground with a force equal to the normal force--else Newton's 3rd law is in trouble!
He'd plow right through the ground!
He does! That's why he sinks down.
The normal force had better be significantly bigger than the impact force, but that still doesn't answer my question. What does the normal force have to do with the force with which the paratrooper hits the ground? The normal force only has to do with how quickly the paratrooper stops, no?
The normal force is the impact force! And, yes, the normal force is what determines how quickly he will stop.

[cookiemonster]

Ah, we're running around with different definitions.

cookiemonster

[KingNothing]

I see...well, I think force normal is the force that the earth applies to the paratrooper...I'm not sure, I never intended to think about it too much...

Anyhoo, this is what I did:

V^2=V^2+2AX
0=16+.4A
-16=.4A
A=-40

d=.5AT^2
.2=20T^2
.01=T^2
T=0.1 s

Change in momentum=F*T
(65)(4)=0.1F
260=0.1F
F=2600 N

Whereas the normal force when he's standing there would be m*a or 637 N.

Now, that also means that if a person who weighed 637 N or 143.3 pounds jumped about

d=4.9(4/9.8)^2

81.6 cm into the air, when he hits the scale, it would show about 585 pounds if the collision was totally elastic (I think thats the right way to say it...assuming he didn't slowly brace himself with bending of the knees). It would hurt to land with straight legs after jumping 81.6 cm into the air.

What do you know, the physics idiot knows how to do something!:P

[ShawnD]

V^2=V^2+2AX
0=16+.4A
-16=.4A
A=-40

d=.5AT^2
.2=20T^2
.01=T^2
T=0.1 s

Change in momentum=F*T
(65)(4)=0.1F
260=0.1F
F=2600 N


You forgot the force of gravity. 2600 would be the answer if the velocities and distances were all horizontal such as a car hitting a wall. The added gravity force is 65 * 9.8 = 637N which would make the answer 3237.

Try it with the energy formula I posted at the top

F_{ground} = \frac{\frac{1}{2}(65)(4^2) + (65)(9.8)(0.2)}{0.2}

F_{ground} = 3237N

Any thoughts or flaws in my reasoning?

[KingNothing]

Perhaps your biggest flaw is your lack of flaws. It may pop other's self esteem balloons. ;)

[ShawnD]

Yeah, sorry about that :wink:

[KingNothing]

Actually, I can think of one potential problem - since the original question was about a paratrooper hitting the ground - wouldn't he reach terminal velocity before hitting the ground. I'm pretty sure a parachuting person would reach terminal velocity. So wouldn't that make his acceleration 0, and not 9.8?

[Janitor]

If he was at terminal velocity, his acceleration was zero immediately before hitting the ground.

His acceleration history upon hitting the ground really depends on the soil mechanics. Imagine the top 8 centimeters is mud, and under that there is 6 centimeters of hard, dry clay, and then sand the rest of the way down. A graph of his acceleration (deceleration, if you prefer) vs. time would be pretty jagged. Probably they expect you to make an assumption of uniform soil composition.