As you move
up to create heavier and heavier nuclei via fusion, the fusions continue
to release energy but in decreasing quantities until the final product
is iron. Thereafter, if you want to fuse nuclei, you must add energy.
So, elements heavier than iron are not produced in stars. Heavier
elements are produced in supernova explosions.

You ask if "…creation of matter from energy…" has ever
been observed. It happens all the time. But that is really the wrong
question because matter and energy are not different things, matter is
simply a form of energy. So, for example, if you fused two ^{32}S
nuclei to a single ^{64}Ge nucleus, the Ge would have more mass
than the two S. Any time you have an endothermic chemical reaction you
end up with more mass than you started with (although almost immesurably
small because chemistry is so inefficient).

*As can be seen in the figure above, the proton-proton cycle requires an
input of 6 protons and ends with one ^{4}He and 2 protons. So the net input
is 4 protons. However, there are numerous outputs—2 positrons, 2
neutrinos, and 2 photons, all of which carry energy away. That is why
comparing the net input with the final product is not meaningful.

Of course there are lots of other ways you could
lift it which would be more efficient if your aim was to tip it over;
for example, you could start pushing horizontally once you got it off
the ground so that the floor would hold up all the weight rather than
half the weight. There is an
old answer very
similar to yours that you might be interested in.

At this point what a physicist normally does is to try to understand the
situation in terms of a simple spring model. If the rope is like a
simple spring,* i.e.* its tension is proportional to its stretch,
you can usually make approximations which would result in the simple
model that *F≈ks* where *k* is a constant. For the
equilibrium situation, then, 20*≈*0.41*k
*or *k≈*49 kg/m. This would then predict that the force
when *s*=1 m would be *F*=49 kg which is too small by more
than a factor of 2.

The previous try indicates that the rope is probably not approximated as
an ideal spring. My last attempt is to try to treat the rope more
explicitly as an ideal spring, not using small approximations used
above. So, looking at one half the rope, I will write* T*=50+*kδ*
where
*δ=*√(*L*^{2}+*s*^{2})-*L
*is the amount by which this half of the rope is stretched relative
to *L*. Note that *k* is not the same as in the
approximation above where* s *was the stretch parameter rather
than *δ. *Calculating *δ *for the at rest
situation where *T*=100 kg, *k*=2270 kg/m. Then,
calculating *T* for the maximum *s* using this value of*
k*, *T*=50+2270x0.129=344 kg which is too large by nearly 60%
compared to the measured value of 220.

I conclude that the rope is poorly approximated as an ideal spring and
that to do any more detailed analysis of this problem would require
measuring *s* as a function of the load by varying the load for
the equilibrium situations.

It is, of course, simple to calculate the speed the weight has just
before the leash goes taught: *v*=√(2*gh*) where is
the distance fallen.

I have deleted
your question about angular momentum—it is off topic.

The
thing to appreciate is that even though the rocket goes straight up, it
will have the same angular velocity *ω *as the earth so its
speed will be *ω*(*L+R*) where *L* is the
length of the rope and *R* is the radius of the earth. The angular
velocity is *ω*=[(2π radians)/(24 hours)]x[(1 hour)/(3600
seconds)]=7.27x10^{-5} s^{-1}. If *L *is just
right, the rocket will assume an orbit like the
geosynchronous communication satellites; this turns out to be if *L*=5.6*R*.

Finally, there are two places on earth where the fictitious forces are
zero and therefore the two objects will move identically—at the
north and south poles.

One more special case. Suppose *
v'*=½*v*_{1}. You should be able to convince yourself that
*m* originally is catching up with the moving frame but then stops
and turns around and ends up where it started from; thereafter it moves
with constant speed *v'*. The total distance it has traveled in the
moving frame is zero and so *W*=0 and therefore Δ*K'=*0.
So the heat is not anywhere evident here even though you would feel it. In
this case you would find *W*=*-*½*mv*_{1}^{2}+½*mv*_{1}^{2}.
Now you can see where that heat is hiding, right?

So, if you want
to most conveniently determine the amount of heat generated, you apply the
work-energy theorem in the frame where the surface is at rest. This is
because to the moving observer the surface is moving as well as *m*. The
moving observer has to correct for that motion by calculating *m*'s motion relative to the surface.
*m*'s motion relative to the second observer is not
really relevant in calculating the heat generated.

Thanks to R. M.
Wood and A. K. Edwards for helpful comments.

The top clock
will see 57 ns more time elapse than the bottom clock. I believe that if
you use the 250 m tall building, things would simply scale and time
differences would be 250/3x10^{8}=8.3x10^{-7} times
smaller.

So the wire
will have a voltage across its ends and will therefore look like a
battery. Similarly, the body will have a voltage of the same polarity (say
positive at the mouth and positive at the end of the wire in your mouth)
but with a much smaller voltage than the wire; this would create a circuit
looking like two batteries, one weak and one strong, which would drive a
current through the weaker battery (your body). So, the idea works if you
move fast enough. But, the earth's magnetic field is really weak and the
speeds would be impossible to achieve without burning everything up in the
atmosphere I would bet. Let's just do a rough calculation. The voltage *
V* is about *V=BLv *where *B* is the field, *L*
is the length, and *v* is the velocity. I will take *L*=1 m,
*v*=18,000 mph≈8000 m/s (low earth orbital speed), and *B*≈5x10^{-5}
T. So the voltage would be less than half a volt! Have I been colloquial
enough for you?

These laws are laws because they
are always found to be true and it is rather pointless to ask "what if".
If energy conservation were not true, then energy could suddenly appear
out of nothing.

Only if there
is some overriding physical law could another be broken, but only within
certain constraints. For example, the Heisenberg uncertainty principle
says that it is impossible to precisely know the energy of a system for a
short enough time, Δ*E*Δ*t*~10^{-34} J⋅s
where Δ*E* is the amount by which you change the energy of a
system and Δ*t* is the time during which the energy is
changed by that amount. For example, 10^{-20} J could appear out
of nothing for as long as 10^{-14} s; after the time had elapsed,
though, it would have to disappear again.

Estimating the force this guy experienced when he hit the ground is a bit
trickier, because what really matters is how quickly he stopped. Keep in
mind that this is only a rough estimate because I do not know the exact
nature of how the ground behaved when he hit it. The main principle is
Newton's second law which may be stated as *F=m*Δ*v*/Δ*t
*where *m* is the mass, Δ*t* is the time to stop, Δ*v=*23
m/s is the change in speed over that time, and *F *is the average force
experienced over Δ*t. *You can see that the shorter the time,
the greater the force; he will be hurt a lot more falling on concrete than on a
pile of
mattresses. I was told that his weight was 156 lb which is *m*=71 kg
and he fell onto about 2" of pine straw; that was
probably over relatively soft earth which would have compressed a couple of
more inches. So let's say he stopped over a distance of about 4"≈0.1 m. We
can estimate the stopping time from the stopping distance by assuming that
the decceleration is constant; without going into details, this results in
the approximate time Δ*t*≈0.01 s. Putting all that into
the equation above for *F*, F≈71x23/0.01=163,000 N≈37,000
lb. This is a very large force, but keep in mind that if he hits flat it
is spread out over his whole body, so we should really think about
pressure; estimating his total area to be about 2 m^{2}, I find
that this results in a pressure of about 82,000 N/m^{2}=12 lb/in^{2}.
That is still a pretty big force but you could certainly endure a force of
12 lb exerted over one square inch of your body pretty easily.

Another possibility is that the victim employed some variation of the
technique parachuters use when hitting the ground, going feet first and
using bending of the knees to lengthen the time of collision. Supposing
that he has about 0.8 m of leg and body bending to apply, his stopping
distance is about eight times as large which would result in in an eight
times smaller average force, about 5,000 lb.

Falling from a third story window (about 32 feet, say) would result in a
speed of about 14 m/s (31 mph) so the force would be reduced by a factor
of a little less than a half.