A skier of mass 60 kg starts at the top of a steep slope with an incline of 40 degrees . The slope has a slight coefficient of k
inetic friction of 0.05 . The slope is L = 100 meters long. At the bottom of the slope, there is an ice ramp (frictionless) that is 10 meters long and angled up at 30 degrees.a) Draw a diagram of the problem.b) How much potential energy does the skier start with?c) How much total energy does the skier start with?d) How much energy is lost to friction on the slope?e) How fast is the skier traveling when she flies off the ramp?f) If there is a large trampoline positioned to catch the skier at the same altitude (y-position) as the beginning of the ramp, where should it be positioned?g) The trampoline behaves like a spring. If the first bounce on the trampoline depresses the trampoline surface by delta y=1m, Without worrying about energy losses, what is the characteristic spring constant of the trampoline?h) Due to losses in the system, the trampoline losses about 2000 Joules every bounce. How many times does the skier bounce? Where does the 1000 Joules on each bounce go? Speculate.
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(a)
The x-rays in the tube are emitted as a result of the collisions of electrons (accelerated through the potential difference applied) on the metal target. Therefore, all the energy of the accelerated electron is converted into energy of the emitted photon:
where the term on the left is the electric potential energy given by the electron, and the term on the right is the energy of the emitted photon, and where:
is the electron's charge
is the potential difference
is the Planck constant
is the speed of light
is the wavelength of the emitted photon
Solving the formula for , we find:
(b) 93300 eV (93.3 keV)
The energy of the emitted photon is given by:
where
h is Planck constant
c is the speed of light
is the wavelength of the photon, calculated previously
Substituting,
Now if we want to convert into electronvolts, we have to divide by the charge of the electron:
(c) The following statements are correct:
The maximum photon energy is just the applied voltage times the electron charge. (1)
The value of the voltage in volts equals the value of the maximum photon energy in electron volts.
In fact, we see that statement (1) corresponds to the equation that we wrote in part (a):
While statement (2) is also true, since in part (b) we found that the photon energy is 93.3 keV, while the voltage was 93.3 kV.