Respuesta :
1) The speed of the skier at the bottom of the valley is 29.7 m/s
2) The speed of the skier at the top of the second hill is 9.9 m/s
3) No, the angle of the hills does not affect the result
Explanation:
1)
We can solve this problem by using the law of conservation of energy. In fact, in absence of friction, the total mechanical energy of the skier is conserved:
[tex]E=U+K[/tex]
where
E is the total mechanical energy
U is the gravitational potential energy
K is the kinetic energy
At the top of the hill, K = 0 since the skier is at rest, so all its energy is potential energy:
[tex]E=U=mgh[/tex] (1)
where
m is the mass of the skier
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
h = 45.0 m is the height of the hill
As the skier descends the hill, the potential energy is converted into kinetic energy. At the bottom, all the mechanical energy has been converted into kinetic energy:
[tex]E=K=\frac{1}{2}mv^2[/tex]
where
v is the speed of the skier at the bottom of the hill
Since the total energy is conserved,
[tex]U=K\\mgh=\frac{1}{2}mv^2[/tex]
And so we find
[tex]v=\sqrt{2gh}=\sqrt{2(9.8)(45)}=29.7 m/s[/tex]
2)
The top of the next hill is located at a height of
h' = 40.0 m
So the total mechanical energy at the top of the second hill is
[tex]E=U'+K' = mgh'+\frac{1}{2}mv'^2[/tex]
where v' is the speed at the top of the second hill.
Since the total mechanical energy must be conserved, we can equate this energy to mechanical energy at the beginning (eq. 1), so we have
[tex]mgh = mgh' + \frac{1}{2}mv'^2[/tex]
and we can now solve for v':
[tex]v=\sqrt{2g(h-h')}=\sqrt{2(9.8)(45-40)}=9.9 m/s[/tex]
3)
As we saw from the previous equations, the angles of the hill does not enter at all the calculations, so it does not affect the value of the speed of the skier.
The reason for this is that the gravitational potential energy of the skier depends only on the height of the hill, h and h', and not from the length of the path along the hill (and so, not on the angle of the hill), and therefore the angle does not enter the calculation.
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(i) The speed at the bottom of the valley is 29.7 m/s
(ii) The speed at the top of the second hill is 9.9 m/s
(iii) The results are independent of the angle of the hills
Since there is no frictional force involved then we can apply the law of conservation of energy, that the total mechanical energy of a system remains conserved under the action of conservative forces like gravitational forces.
E = KE + PE
where KE is kinetic energy and PE is potential energy
At the top of the hill, KE = 0 since the skier is at rest, so all its energy is potential energy:
PE = mgh
where
m is the mass of the skier
g is the acceleration of gravity
h = 45.0 m is the height of the hill
At the bottom, there is only the kinetic energy:
[tex]KE = \frac{1}{2}mv^2[/tex]
where
v is the speed of the skier at the bottom of the hill
Since the total energy is conserved, then:
KE = PE
[tex]\frac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 45}m/s\\ \\v=29.7m/s[/tex]
The height of the second hill is h' = 40.0 m
Applying the conservation of energy, the kinetic energy at the bottom will be converted to potential energy and remaining kinetic energy for the second hill:
let the speed of the skier at the top of second hill be v', then
[tex]\frac{1}{2}mv^2=mgh'+\frac{1}{2}mv'^2\\\\ mgh=mgh'+\frac{1}{2} mv'^2\\\\v'=\sqrt{2g(h-h')}\\ \\v'=\sqrt{2\times9.8\times5} m/s\\\\v'=9.9m/s[/tex]
We can clearly see in the above results that the results are independent of the angle of the hills.
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