Respuesta :
The image is missing, I attach it in the answer.
part a) The roller coaster starts from [tex]h_1 = 25 m[/tex] from rest, this means that its initial mechanical energy is only potential energy:
[tex]E= mgh_1[/tex]
When it reaches the bottom point at altitude [tex]h_2 = 3.0 m[/tex], the roller coaster also has kinetic energy, therefore the total mechanical energy at this point will be
[tex]E=mgh_2 + \frac{1}{2} mv_2 ^2[/tex]
The energy must be conserved, so we can write
[tex]mgh_1 = mgh_2 + \frac{1}{2}mv_2^2 [/tex]
from which we find the speed of the roller coaster at the bottom point
[tex]v_2 = \sqrt{2g(h_1 - h_2 )}= \sqrt{2 (9.81 m/s^2)(25 m-3m)} =20.8 m/s [/tex]
part b) The top of the loop at point B is at altitude [tex]h_B = 12.0 m[/tex] from the ground, so we can use the same producedure of part a) to find the speed at point B:
[tex]mgh_1 = mgh_B + \frac{1}{2}m v_B^2[/tex]
from which we find
[tex]v_B = \sqrt{2g(h_1 - h_B)}= \sqrt{2(9.81 m/s^2)(25 m-12 m)} =16.0 m/s [/tex]
part c) Now we have a part of the track where there is frictional force acting on the roller coaster: this means that the mechanical energy is no longer conserved in that part of the track.
In particular, the frictional force against the roller coaster is F=625 N, and it acts for a distance of d=9.4 m, therefore the work done by the friction (which corresponds to the energy lost by the car) is
[tex]W=\Delta E=Fd=(625 N)(9.4 m)=5875 J[/tex]
At point B, the car should have a speed of [tex]v_B=7.7 m/s[/tex]. The altitude at that point is [tex]h_B = 12.0 m[/tex], therefore the mechanical energy at point B is
[tex]E_B = mgh_b + \frac{1}{2}mv_B^2 = (50 kg)(9.81 m/s^2)(12m)+ \frac{1}{2}(50kg)(7.7m/s)^2= [/tex]
[tex]=7368 J[/tex]
Therefore, the car should have at point A an energy of
[tex]E_A = E_B + \Delta E= 7368 J + 5875 J =13243 J[/tex]
The gravitational potential energy of the car at point A is
[tex]U_A = mgh_A=(50 kg)(9.81 m/s^2)(25 m)=12263 J[/tex]
So the kinetic energy must be
[tex]K_A = E_A - U_A=13243 J - 12263 J =980 J[/tex]
and so the velocity must be:
[tex]v_A = \sqrt{ \frac{2K_A}{m} }= \sqrt{ \frac{2(980 J)}{50 kg} }=6.3 m/s [/tex]
part a) The roller coaster starts from [tex]h_1 = 25 m[/tex] from rest, this means that its initial mechanical energy is only potential energy:
[tex]E= mgh_1[/tex]
When it reaches the bottom point at altitude [tex]h_2 = 3.0 m[/tex], the roller coaster also has kinetic energy, therefore the total mechanical energy at this point will be
[tex]E=mgh_2 + \frac{1}{2} mv_2 ^2[/tex]
The energy must be conserved, so we can write
[tex]mgh_1 = mgh_2 + \frac{1}{2}mv_2^2 [/tex]
from which we find the speed of the roller coaster at the bottom point
[tex]v_2 = \sqrt{2g(h_1 - h_2 )}= \sqrt{2 (9.81 m/s^2)(25 m-3m)} =20.8 m/s [/tex]
part b) The top of the loop at point B is at altitude [tex]h_B = 12.0 m[/tex] from the ground, so we can use the same producedure of part a) to find the speed at point B:
[tex]mgh_1 = mgh_B + \frac{1}{2}m v_B^2[/tex]
from which we find
[tex]v_B = \sqrt{2g(h_1 - h_B)}= \sqrt{2(9.81 m/s^2)(25 m-12 m)} =16.0 m/s [/tex]
part c) Now we have a part of the track where there is frictional force acting on the roller coaster: this means that the mechanical energy is no longer conserved in that part of the track.
In particular, the frictional force against the roller coaster is F=625 N, and it acts for a distance of d=9.4 m, therefore the work done by the friction (which corresponds to the energy lost by the car) is
[tex]W=\Delta E=Fd=(625 N)(9.4 m)=5875 J[/tex]
At point B, the car should have a speed of [tex]v_B=7.7 m/s[/tex]. The altitude at that point is [tex]h_B = 12.0 m[/tex], therefore the mechanical energy at point B is
[tex]E_B = mgh_b + \frac{1}{2}mv_B^2 = (50 kg)(9.81 m/s^2)(12m)+ \frac{1}{2}(50kg)(7.7m/s)^2= [/tex]
[tex]=7368 J[/tex]
Therefore, the car should have at point A an energy of
[tex]E_A = E_B + \Delta E= 7368 J + 5875 J =13243 J[/tex]
The gravitational potential energy of the car at point A is
[tex]U_A = mgh_A=(50 kg)(9.81 m/s^2)(25 m)=12263 J[/tex]
So the kinetic energy must be
[tex]K_A = E_A - U_A=13243 J - 12263 J =980 J[/tex]
and so the velocity must be:
[tex]v_A = \sqrt{ \frac{2K_A}{m} }= \sqrt{ \frac{2(980 J)}{50 kg} }=6.3 m/s [/tex]
a) The speed of the roller coaster as it reaches the [tex]3.0\,{\text{m}}[/tex] will be [tex]\boxed{20.8\text{ m/s}}[/tex].
b) The speed of the roller coaster as it reaches the [tex]\boxed{12.0\text{ m}}[/tex] will be [tex]16.0\text{ m/s}[/tex].
c) The speed of the car at the point A should be [tex]\boxed{6.3\text{ m/s}}[/tex] to reach the top point of the loop.
Further Explanation:
Part (a)
Since the roller coaster starts from the top point of the loop which is at a height of [tex]25\text{ m}[/tex] from the ground. Therefore, at this point, the roller coaster has the mechanical energy stored in the form of potential energy due to the position of the roller coaster.
As the roller coaster starts moving, the potential energy of the roller coaster starts to convert into the kinetic energy as the roller coaster gains speed down the road.
The potential energy of the roller coaster as it starts from top point A is:
[tex]\boxed{E = mg{h_A}}[/tex]
Here, [tex]{h_A}[/tex] is the height of the top point A from the ground.
The mechanical energy of the roller coaster as it reaches to the point [tex]3.0\text{ m}[/tex] above the ground is:
[tex]E=mg{h_2}+\dfrac{1}{2}m{v^2}[/tex]
The energy of the roller coaster remains conserved throughout its motion. So, on equating the energy of the roller coaster at two positions, we can obtain the speed of the roller coaster at the point [tex]3.0\text{ m}[/tex] above the ground.
[tex]\begin{aligned}mg{h_A}&=mg{h_2}+\frac{1}{2}m{v^2}\\v&=\sqrt{2g({{h_A}-{h_2}})}\\\end{aligned}[/tex]
Substitute the values.
[tex]\begin{aligned}v& = \sqrt{2({9.81\text{ m/s}^2)({25\,{\text{m}}-3\,{\text{m}}})}\\&=\sqrt{({19.62})( {22})}{{\text{ m/s}}\\&=20.8\text{ m/s}\\\end{aligned}[/tex]
Therefore, the speed of the roller coaster as it reaches the point [tex]3.0\text{ m}[/tex] above the ground is [tex]20.8\text{ m/s}[/tex].
Part (b)
The roller coaster now reaches the top of the loop at the point B which is at a height of [tex]12.0\text{ m}[/tex] above the ground. The energy of the roller coaster remains conserved.
[tex]\begin{aligned}mg{h_1}&=mg{h_B}+\frac{1}{2}m{v_B^2}\\v_B&=\sqrt{2g({{h_1}-{h_B}})}\\\end{aligned}[/tex]
Substitute the values.
[tex]\begin{aligned}v_B&=\sqrt {2({9.81\text{ m/s}^2)({25\text{ m}-12\text{ m}})}\\&=\sqrt{({19.62})({13})}\text{ m/s}\\&=16.0\text{ m/s}\\\end{aligned}[/tex]
Therefore, the speed of the roller coaster as it reaches the top point of the loop at [tex]12.0\text{ m}[/tex] above the ground is [tex]16.0\text{ m/s}[/tex].
Part (c)
The roller coaster needs to reach the top point of the loop with the minimum speed of [tex]7.7\text{ m/s}[/tex]. The mechanical energy possessed by the [tex]50\text{ kg}[/tex] roller coaster as it reaches the top point of the loop should be the sum of the potential energy and the kinetic energy of the roller coaster.
The energy possessed by the roller coaster at the top point of the loop (point B) is:
[tex]{E_B}=mg{h_B}+\dfrac{1}{2}mv_B^2[/tex]
Substitute the values.
[tex]\begin{aligned}{E_B}& = \left[ {({50\text{ kg}})({9.81\text{ m/s}^2)({12\text{ m}})}\right]+\left[ {\frac{1}{2}({50\text{ kg}}){{({7.7\text{ m/s})}^2}}\right]\\&=5886 + 1482\text{ J}}\\ &=7{\text{368}}\,{\text{J}}\\\end{aligned}[/tex]
We can get the work done by the friction force as:
[tex]\begin{aligned}{W_f}&=F\cdot s\\&=({625\text{ N}})\cdot({9.4\text{ m}})\\&=5875\text{ J}}\\\end{aligned}[/tex]
Therefore, the roller coaster needs to have the amount of energy at the point A which is sufficient to overcome the friction and also have the sufficient energy to reach the point B with required amount of mechanical energy.
The energy of the roller coaster at the point A should be:
[tex]\begin{aligned}{E_A}&={W_f}+{E_B}\\&=5875\text{ J}+7368\text{ J}\\&=13243\text{ J}}\\\end{aligned}[/tex]
The total energy possessed by the roller coaster should be the sum of the kinetic energy and the potential energy.
The total energy of the roller coaster at point A is:
[tex]\begin{aligned}{E_A}&=mg{h_A}+\frac{1}{2}mv_A^2\\13243\text{ J}&=\left[{({50\text{ kg}}})({9.81\text{ m/s}^2)({25\text{ m}})} \right]+\left[{\frac{1}{2}( {50\text{ kg}})v_A^2}\right]\\{v_A}&=\sqrt{\frac{{2({980})}}{{50}}}\\&=6.3\text{ m/s}\\\end{aligned}[/tex]
Thus, the roller coaster must have a speed of [tex]6.3\text{ m/s}[/tex] at point A so that it reaches the point B without leaving the contact with the surface.
Learn More:
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Answer Details:
Grade: College
Subject: Physics
Chapter: Conservation of energy
Keywords:
Roller coaster, loop, point A, point B, energy, mechanical energy, potential, kinetic, 25 m, 3 m, 12 m, conserved, frictionless, friction force, 50 kg car, speed of roller coaster, 625 N.
