Respuesta :
1) The speed of the roller coaster at ground level is 24.2 m/s
2) The speed of the roller coaster at a point 25.0 m above the ground is 9.9 m/s
3) The speed of the roller coaster at a point 12.0 m above the ground is 18.8 m/s
Explanation:
1)
We can solve this problem by using the law of conservation of energy.
In fact, the total mechanical energy of the roller coaster at any point along the track must be conserved, so it is constant:
[tex]E=K+U=const.[/tex]
where
K is the kinetic energy
U is the gravitational potential energy
At point A, the initial point, the roller coaster is still at rest, so its mechanical energy is all gravitational potential energy:
[tex]E=U_A = mgh_A[/tex] (1)
where
m is the mass
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
[tex]h_A = 30 m[/tex] is the initial height
When the roller coaster reaches the ground level, h = 0, so all the mechanical energy has been converted into kinetic energy:
[tex]E=K_B = \frac{1}{2}mv_B^2[/tex] (2)
where
[tex]v_B[/tex] is the speed of the cart at point B (the ground level)
Equating (1) and (2), we can find the speed of the cart at ground level:
[tex]mgh_A = \frac{1}{2}mv_B^2\\v_B = \sqrt{2gh_A}=\sqrt{2(9.8)(30.0)}=24.2 m/s[/tex]
2)
The mechanical energy at point C (the point at 25.0 m above the ground) is sum of potential energy and kinetic energy:
[tex]E=U_C + K_C = mgh_C + \frac{1}{2}mv_C^2[/tex] (3)
where
[tex]h_C = 25.0 m[/tex] is the height above the ground
[tex]v_C[/tex] is the speed at point C
We already know that the total mechanical energy is
[tex]E=mgh_A[/tex] (4)
where [tex]h_A = 30.0 m[/tex] is the initial height
Since the total energy is conserved, by equating (3) and (4) we find the speed at point C:
[tex]mgh_A = mgh_C + \frac{1}{2}mv_C^2\\v_C = \sqrt{2g(h_A-h_C)}=\sqrt{2(9.8)(30.0-25.0)}=9.9 m/s[/tex]
3)
This part is similar to part 2): the mechanical energy at point D (the point at 12.0 m above the ground) is sum of potential energy and kinetic energy, so
[tex]E=U_D + K_D = mgh_D + \frac{1}{2}mv_D^2[/tex] (5)
where
[tex]h_D = 12.0 m[/tex] is the height above the ground
[tex]v_D[/tex] is the speed at point D
We know that the total mechanical energy is
[tex]E=mgh_A[/tex] (6)
where [tex]h_A = 30.0 m[/tex] is the initial height
Since the total energy is conserved, by equating (5) and (6) we find the speed at point D:
[tex]mgh_A = mgh_D + \frac{1}{2}mv_D^2\\v_D = \sqrt{2g(h_A-h_D)}=\sqrt{2(9.8)(30.0-12.0)}=18.8 m/s[/tex]
Learn more about kinetic energy and potential energy:
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