How Can Force and Time Determine Velocity in Physics Problems?

[Ricaoma]

Homework Statement

The graph below shows the force as a function of time when a man jumps on a bathroom scale.

1) Determine the mans weight (mass, kg) using the graph.
2) Describe what happens in the time intervals A, B, C and D.
3) Determine the velocity of the man as he leaves the scale (the velocity at the beginning of his jump).

Unnecessary high resolution picture and since I have no idea how to limit the size here, I'll just link the graph instead.
http://img146.imageshack.us/img146/8286/grafq.jpg"

Homework Equations

No specific equations are given, but I'd assume the equations regarding potential energy, kinetic energy, velocity, and force would be relevant.

Though, this far I've only used
F = m×g \Leftrightarrow m = \frac{F}{g}
to calculate an approximated mass of the man.

The Attempt at a Solution

1)
F \approx 700 N
g \approx 9.82 m/s²
m \approx \frac{700}{9.82} \approx 71.3 kg

2)
A The man starts his preparations for the jump by bending his knees. (Why does this make the force lower? Is it because of his center of mass changing?
B He re-straightens his legs, thrusting himself upwards. This makes his feet push down on the scale, making it register a higher force.
C The man is in the air and is therefore not touching the scale.
D The man lands on the scale with greater force because of the fall (related to gravity).

3)
This is were I'm stuck. At first I though about using the time he was in the air to calculate this, but I realized I didn't really know how.
Then, I changed my plan to using the force with which he thrusted himself upwards with, and using the number F = 1800N with the formula
F = m×A \Leftrightarrow A = \frac{F}{m}
to later be able to calculate his velocity using
V = A×\Delta T
until I realized that I would have to assume a time and that this assumed time would be completely responsible for the result, which seems a bit to odd for an answer.

 

[Fewmet]

Welcome to Physics Forums.

Your answer to #1 looks good.

I think your reasoning is a little lax on the landing. The "force of the fall" is pretty vague. Think about this in terms of the third law of motion. The scale reads teh fore pushing down in it, and therefore the force it pushes up with. Why does it push up with less force on the standing man than on the landing man?

(Similar reasoning with answer your question about why the scale reads less during squatting).

On #3: do you know impulse-momentum relationship? That'll work well here

 

[Ricaoma]

Thanks.

And yes, my arguments for why these things are happening on #2 are pretty vague. Thanks for the help, really appreciate it.

I assume this would be caused because of the scale having to push the man upward with a higher force to stop his fall, as opposed to simply holding him in place?
When it comes to the squatting part, is it because of a lower potential energy? Or is it something completely different?

And about #3; actually, we haven't talked about impulse-momentum relationships in class, but I did some google-searching to get an idea what it was all about. I'm still not sure if I know enough about it to actually use it to calculate the speed without some pretty detailed guidance, though. The things we've talked about in class recently is basicly energy (kinetic and potential) and distance-time/velocity-time/acceleration-time graphs.

 

[Fewmet]

Thanks.

And yes, my arguments for why these things are happening on #2 are pretty vague. Thanks for the help, really appreciate it.

I assume this would be caused because of the scale having to push the man upward with a higher force to stop his fall, as opposed to simply holding him in place?
When it comes to the squatting part, is it because of a lower potential energy? Or is it something completely different?

That's correct for the landing.

About jumping, focus on what the man does to squat. It is not obvious, but when you squat you lift your legs up (in freefall it is much clearer). SO if the man lifts his legs up from the scale, the scale...

And about #3; actually, we haven't talked about impulse-momentum relationships in class, but I did some google-searching to get an idea what it was all about. I'm still not sure if I know enough about it to actually use it to calculate the speed without some pretty detailed guidance, though. The things we've talked about in class recently is basicly energy (kinetic and potential) and distance-time/velocity-time/acceleration-time graphs.

OK, thy this:
F = ma
F= m∆v/∆t
where F is the stooping force (the scale reading), you know m and ∆t is the stopping time, which can be read off the graph.

Note that you will have to be careful about what force was acting during that time. The graph shows a pretty linear change from maximum to minimum, so the average force should be a good approximation.

(Looking back at you original post, I see the method I just sketched out is equivalent to where you started. I simply pointed out how to get the time and to take care with choosing the force.)

You could try to apply the principles of energy to this, but that would require estimating the distance over which the man stops.

 

[Ricaoma]

That's correct for the landing.
OK, thy this:
F = ma
F= m∆v/∆t
where F is the stooping force (the scale reading), you know m and ∆t is the stopping time, which can be read off the graph.

Note that you will have to be careful about what force was acting during that time. The graph shows a pretty linear change from maximum to minimum, so the average force should be a good approximation.

(Looking back at you original post, I see the method I just sketched out is equivalent to where you started. I simply pointed out how to get the time and to take care with choosing the force.)

You could try to apply the principles of energy to this, but that would require estimating the distance over which the man stops.

Let me see if I got this right (English isn't my primary language, nor the language in which my physics education is held, so I might've misunderstood some parts of this last post since I'm not too sure about how to solve this third question to begin with)

F= \frac{m × ∆v}{∆t} \Leftrightarrow ∆v = \frac{F × ∆t}{m}

where
m = 71.3 kg
F_avg = 1000 N
∆t = 5.5-5.3 = 0.2 s

∆v = \frac{1000 × 0.2}{71.3} \approx 2.8 m/s

or am I using the wrong numbers here?

 

[Fewmet]

That looks good.

As a check, think about a body launching vertically at 2.8 m/s with a=-9.8 m/s^2. It then lands with -2.8 m/s, and the time that takes is about 0.57 seconds. That matches pretty well with the time the graph show the man being in the air. (I'd expect it to be a little less that 0.57 seconds because of his legs bending).

 

[Ricaoma]

That looks good.

As a check, think about a body launching vertically at 2.8 m/s with a=-9.8 m/s^2. It then lands with -2.8 m/s, and the time that takes is about 0.57 seconds. That matches pretty well with the time the graph show the man being in the air. (I'd expect it to be a little less that 0.57 seconds because of his legs bending).

Awesome, thanks a lot for all the help! I really appreciate it.

Edit: Also, what's the connection/reason for using the F= m∆v/∆t formula/equation? Any way to connect it to any of the "basic" formulas for the things I mentioned earlier that we had talked about in class (kinetic and potential energy and distance-time/velocity-time/acceleration-time graphs)? It would help to memorize and describe the method if there was actually some connection, otherwise I'll simply have to study this method and try to remember it as a separate method.

 

[Fewmet]

Also, what's the connection/reason for using the F= m∆v/∆t formula/equation? Any way to connect it to any of the "basic" formulas for the things I mentioned earlier that we had talked about in class (kinetic and potential energy and distance-time/velocity-time/acceleration-time graphs)? It would help to memorize and describe the method if there was actually some connection, otherwise I'll simply have to study this method and try to remember it as a separate method.

I suppose it is a matter of style. It is a slight variation on the impulse/momentum relationship (which you haven't studied yet).

Consider this: when you apply a net force to a body it accelerates. If you want to know how fast it ends up going, you could look at over what distance you applied the force. In that case you are looking at word and (as you have studied) the work you do causes a change in energy. You could say
F∆d = ∆(1/2 mv²)

Alternatively, you could find the change in velocity by looking at the time the force was applied. The quantity F∆t is called impulse. When you apply an impulse you change momentum (which mathematically is ∆(mv). So
F∆t = ∆(mv) = m∆v for constant m. Dividing both sides by ∆t gives the relationship in question.