Forces again *sigh* me , this is not homework

[holezch]

forces again.. *sigh* please help me , this is not homework!

1. Homework Statement
a man of mass 80 kg ( weight 176 lb), jumps down to a concrete patio from a window ledge only 0.50 m above ground. He neglects to bend his knees on landing, so that his motion is arrested in a distance of about 2.0 cm.

a) what is the average acceleration of the man from the time his feet first touch the patio to the time he is brought fully to rest?

b) with what average force does this jump jar his bone structure?



2. Homework Equations

f = ma, 3rd law



3. The Attempt at a Solution
I really don't know what to do at all...
if a man falls down and accelerates a little bit forward after the fall, will the same force from falling transfer to that small time he is moving forward?

and how will an acceleration be affected if something is falling from the sky and then bounces off the ground? touching the ground will make it go slower wouldn't it

please help me, this is not homework.. I'm trying to learn physics out of some book and it feels impossible

 

[holezch]

the force of the man falling is not just weigh is it? the jumping will have some kind of force? so, his force will be F = 176 + ma

 

[holezch]

okay, I realized a couple of things.. when the man hits the ground, the force component is canceled out by the ground. so I'm wondering what keeps a thing in motion after it has hit the ground? why does a coin bounce when dropped to the ground? I thought that it was because the velocity accumulated during the fall becomes large enough, when the force cancels out, you still need something that can set the velocity to zero. you can' t have 100 m/s to 0 instantaneously (I don't think so ).. but then why would crouching as you fall do exactly that?

please help me I am desperate

 

[BlueEight]

I can only help you with part a, I'm not so sure about b.

Given:
\Delta{y}=0.5m
\bar{a}=\frac{\Delta{V}}{\Delta{t}}

To find the man's speed at first impact, use the kinematic equations for uniform acceleration:
{V_{imp}}^2={V_o}^2+2g\Delta{y}
Thus,
(1){V_{imp}}=\sqrt{{V_o}^2+2g\Delta{y}}

We now need to find the time it takes the man to decelerate after hitting the ground. Use:
{V}^2={V_{imp}}^2+2a\Delta{y} where final velocity V=0
Which translates to
a=\frac{V^2-{V_{imp}}^2}{2\Delta{y}}

Using this acceleration, find the time for him to decelerate from V_imp to 0 m/s. Another equation is:
V=V_{imp} + at
which becomes:
(2)t=\frac{V-V_{imp}}{a}

Finally, use the values obtained in (1) and (2) and divide them to obtain the average acceleration, which is change in velocity divided by change in time.

Phew! First time using LaTeX. Hope this makes sense!

 

[holezch]

thank you sooooo muchh I will read over whatyou have said

 

[holezch]

I can only help you with part a, I'm not so sure about b.

Given:
\Delta{y}=0.5m
\bar{a}=\frac{\Delta{V}}{\Delta{t}}

To find the man's speed at first impact, use the kinematic equations for uniform acceleration:
{V_{imp}}^2={V_o}^2+2g\Delta{y}
Thus,
(1){V_{imp}}=\sqrt{{V_o}^2+2g\Delta{y}}

We now need to find the time it takes the man to decelerate after hitting the ground. Use:
{V}^2={V_{imp}}^2+2a\Delta{y} where final velocity V=0
Which translates to
a=\frac{V^2-{V_{imp}}^2}{2\Delta{y}}

Using this acceleration, find the time for him to decelerate from V_imp to 0 m/s. Another equation is:
V=V_{imp} + at
which becomes:
(2)t=\frac{V-V_{imp}}{a}

Finally, use the values obtained in (1) and (2) and divide them to obtain the average acceleration, which is change in velocity divided by change in time.

Phew! First time using LaTeX. Hope this makes sense!

will you get a set number if you don't know the velocities?

 

[BlueEight]

You know all the "velocities" that you'll need. Granted, you need to infer it from the problem, but it's there.

 

[holezch]

thanks, so to solve for Vimp, set V = 0 for your "a" equation and then solve for Vimp?

 

[BlueEight]

Yes. (If by V you mean V_o)

 

[holezch]

Yes. (If by V you mean V_o)

no, I meant V, to solve for the time it takes to go from Vimpact to V, V = 0
we don't know Vo

 

[BlueEight]

What? To get V_imp, we use
<br /> {V_{imp}}=\sqrt{{V_o}^2+2g\Delta{y}}<br />

V_o means initial velocity, aka his velocity when he first walks off/jumps off the ledge.

 

[holezch]

What? To get V_imp, we use
<br /> {V_{imp}}=\sqrt{{V_o}^2+2g\Delta{y}}<br />

V_o means initial velocity, aka his velocity when he first walks off/jumps off the ledge.

yes, but what is Vimpact? we can't have a number if we don't know Vo? thanks

 

[BlueEight]

We DO know V_o. It's his INITIAL velocity when he starts falling. Imagine that he's a particle-like object that you're holding. What is his speed right when you drop the object?

 

[holezch]

We DO know V_o. It's his INITIAL velocity when he starts falling. Imagine that he's a particle-like object that you're holding. What is his speed right when you drop the object?

he jumps though, so we don't know his initial velocity? it isn't like he just walks off the thing
by the way, thanks for replying to me and everything..I really appreciate it . I'm going to bed now, I'll check on the thread tomorrow morning :)

also, I think we can solve for Vimp in the 'a' equation, we don't even need Vo

 

[BlueEight]

In these problems, if his initial velocity isn't give, assume he just walks off.

 

[holezch]

In these problems, if his initial velocity isn't give, assume he just walks off.

ohh okay, but I don;t think we even need Vo, is it right? I think we can find Vimp from just the 'a' equation

 

[BlueEight]

You're referring to this, right?
<br /> a=\frac{V^2-{V_{imp}}^2}{2\Delta{y}}<br />

Also, no. That only works for after he hits the ground. Read through my explanation again.

 

[cepheid]

he jumps though, so we don't know his initial velocity? it isn't like he just walks off the thing

Yeah, that is basically what we are assuming here (i.e. he somehow just starts falling from a height of 0.5 m).

BlueEight, it doesn't seem strange to you that your solution assumes constant acceleration after impact? Your steps basically amount to:

1. Assume constant acceleration, and calculate it based on the distance over which it occurs and the change in velocity.

2. Use the calculated acceleration to find the time over which it occurred.

3. Divide the change in velocity by the time to get the acceleration.

If you are allowed to assume a is constant, then steps 2 and 3 are redundant and circular. However, I suspect that assuming the acceleration is constant really makes no sense, because upon impact the force is probably going to vary with time in some complicated way that we have no way of figuring out, which is why we can only speak of the *average* force (and acceleration), which is why that's what the problem asks for.

 

[BlueEight]

From my limited knowledge, normal force is what slows down an object after impact. Therefore, according to Newton's Second Law, Normal force N = mg + ma. However, I do not know how the acceleration will be affected after impact, and thus declare this problem unsolvable, having only gone through a quarter of AP Physics.

But really, we need to somehow find a(t) after impact.

 

[cepheid]

From my limited knowledge, normal force is what slows down an object after impact. Therefore, according to Newton's Second Law, Normal force N = mg + ma. However, I do not know how the acceleration will be affected after impact, and thus declare this problem unsolvable, having only gone through a quarter of AP Physics.

But really, we need to somehow find a(t) after impact.

There is no way to find F(t). Force is the rate of change of momentum, and after impact that is going to depend on all sorts of complicated things like how the object and the ground deform upon impact etc etc. Remember that all contact forces are fundamentally electromagnetic in nature (when you examine things at the scale of the atoms making up the matter in the the two bodies that are in contact). So there is a whole bunch of physics that we are not prepared to really investigate and are basically just glossing over.

For part b, it seems easy enough to calculate the total work done, and from that, to determine the average force, which equals (total work done)/distance. This would be an average over distance, not time. From that, you could deduce an average acceleration (averaged over distance travelled), but that doesn't seem to be what the problem is asking for, especially since it asks for acceleration *first* (in part a). I'm not sure how to solve part a.

 

[holezch]

interesting, I didn't know that it would be this complicated.. if we were allowed to assume that the acceleration is constant after hitting the ground, would BlueEight's answer be okay? I see what you mean by the redundancy though

thanks a lot guys!

 

[Jasso]

All you really need to know for this problem is two equations:


Eq. 1 a_{avg}=\frac{V_f^2-V_i^2}{2(h_f-h_i)}

Eq. 2 f_{avg}=ma_{avg}

Where V_i is the initial velocity, V_f is the final velocity, h_i is the starting height, and h_f is the final height.

And then separate the problem out into two situations, 1) when the person is falling and 2) when the person is stopping.

1) Rearrange Eq. 1 to find the velocity the person reaches when he gets to the patio and set a_{avg}=g=-9.8 m/s^2 , V_i=0 m/s , and h_f-h_i=(0 m)-(0.5 m)= -0.5 m:

V = \sqrt{2g(h_f-h_i)} = \sqrt{2(-9.8 m/s^2)(-0.5 m)}

2) V is the velocity that the person hits the patio with, so use Eq. 1 with V this time being the initial velocity V_i , h_f-h_i= 2 cm= 0.02 m , and since it states that he stops V_f=0 m/s:

a_{avg}=\frac{V^2}{2(0.02 m)}

Use the a_{avg} from that and plug it into f_{avg}=ma_{avg} to find the average force of the patio.

 

[Jasso]

if we were allowed to assume that the acceleration is constant after hitting the ground, would BlueEight's answer be okay?

I think cepheid's point was the roundabout way of using the time of acceleration to find the acceleration, which would only work if the acceleration were constant. The problem does not assume the acceleration is constant, instead it asks for the average acceleration and force.

 

[holezch]

thanks a lot, my understanding of this question has increased A LOT. I haven't read fully your solutions, I picked the starting point so I can see if I can solve it with some hints.. t hanks a lot!

 

[holezch]

hi , I used your method.. I got that acc(avg) is 2,401... how is that even possible?

 

[Jasso]

hi , I used your method.. I got that acc(avg) is 2,401... how is that even possible?

You either squared V twice or forgot that its the square root of 2g(h_f-h_i).

If you plug the V from the first part into the V from the second part you get

a_{avg}=\frac{\sqrt{2(-9.8 m/s^2)(-0.5 m)}^2}{2(0.02 m)}

which simplifies into

a_{avg}=(9.8 m/s^2)\frac{(0.5 m)}{(0.02 m)}.