ANSWER
The number of calories (actually kcal) in a Mars Bar is 294 kcal,
according to the World Cancer Research Fund. This is a measure of
the chemical energy it releases when burnt. One kcal = 4184 J, so a
Mars Bar contains 1.23 MJ. If running up a flight of stairs gains
2352 J of PE, then you'd have to repeat the run 520 times to use up
the energy in the Mars Bar.
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ANSWER1
We've assumed that all the energy in the Mars Bar is converted to the
gain in PE in running upstairs. In fact, no energy conversion is 100%
efficient and some energy goes into working against friction in the
body joints etc. So some energy is converted into heat rather than
into PE. You can see that this must be true because it takes energy to
run on the flat, where there is no gain in PE. Infact, when you run
on the flat, you do work (and so use energy) raising & lowering your
centre of gravity as you run.
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ANSWER2
If you walk up the stairs, you gain the same amount of PE, since that
depends only on your mass, g, and the vertical height of the stairs.
You take a longer time to reach the top though, and so your rate of
using energy (or your rate of doing work), is lower than if you ran up
the stairs. Your rate of using energy is your power
output. Running requires high power for a short time; walking requires
lower power for a longer time.
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If you walk, why aren't you as hot and sweaty at the top?
ANSWER2A
Most of our energy intake is used in keeping our bodies warm.
2500 kcal/day = 2500 x 4184 J/kcal / 24 x 3600 = 120 W
To see if we could use 120 W doing physical work, take walking up
stairs as an example again. If it takes 20s to walk up a flight of
stairs, the power is 2352 J /20 s = 118 W, divided by the bodies
efficiency (eg assume 30%), giving 400 W. So to use up the energy in
our food, we would have to continuously walk up stairs for about a
quarter of each day.
To make a comparison, one horse-power is 750 W, and cars have typically
100 horse-power.
Click here to return. ANSWER3
At high power, the sudden creation of heat raises the body
temperature. This heat has to be lost to maintain the constant body
temperature necessary for the body to function. At lower power, the
heat can be lost through normal body cooling processes while walking.
Click here to return. ANSWER4
The PE we've given to XMM is enough for it to reach the height of its
orbit, but it will have zero velocity at that point. Importantly, we
need to stop XMM falling straight back down to Earth ! As well as
getting it to the height of the orbit we need to give it a velocity to
keep it in orbit (actually, an acceleration to match the Earths
gravitational attraction). Remember that the gain in PE in the Earths
(radial) gravitational field depends only on the start and end
heights, and is equal to the work done in moving between them.
The extra energy we need to give XMM is kinetic energy KE = 1/2
mv2. Since Fgrav = GMm/r2 =
mv2/r, KE = 1/2 GMm/r. The gain in PE we've given to XMM
is mg
The total energy of XMM in orbit, E, is the sum of the KE and the PE
in orbit (ie as measured from the centre of the Earth, not just the
gain in PE in the launch): (remember g= -GM/r2)
With zero PE defined to be at infinity, the total energy in a bound orbit
is negative.
Control a
simulation
of the different orbits (and non-orbits!) produced by different launch
velocities (uses Java).
In practice there are also some extra complications in calculating the
launch energy: g depends on distance from the centre of the Earth (g=
-GM/r2), and so is not a constant if the orbit is high compared to
the radius of the Earth.
The mass of the rocket also decreases significantly with time as fuel is burnt.
A diagram showing the orbit of XMM
(it is elliptical, not circular)
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h = m (GM/r2) h. If h is approximately equal to r/2
(because the height of the orbit is about the same as the Earths
radius), then the KE we need to give to XMM has about the same value
as the PE gain we've already given.
E = KE + PE = 1/2 mv2 + mgr = 1/2 GMm/r - m (GM/r2) r
= 1/2 GMm/r - GMm/r = -1/2 GMm/r
