4. Gravitational Potential Energy. In TWICE TV2, Dahyun (mass = 48.9 kg), Jeongyeon (mass = 49.1 kg), and Chaeyoung (mass = 46 kg) were asked to ride an amusement park roller coaster. They were testing this roller coaster You are testing a new amusement park roller coaster with a car of mass 120 kg (disregard the mass of other staff, we are interested with TWICE only). One part of the track is a vertical loop with radius 12.0 m. At the bottom of the loop (point A) the car has speed 25.0 m/s and at the top of the loop (point B) it has speed 8.0 m's. As the car (with our TWICE idols) rolls from point A to point B, (a) how much work is done by friction? Using work energy theorem, (b) what is the gravitational potential energy? (c) what is the speed of the roller coaster car? (d) bonus: illustrate the problem. (5 %3D
4. Gravitational Potential Energy. In TWICE TV2, Dahyun (mass = 48.9 kg), Jeongyeon (mass = 49.1 kg), and Chaeyoung (mass = 46 kg) were asked to ride an amusement park roller coaster. They were testing this roller coaster You are testing a new amusement park roller coaster with a car of mass 120 kg (disregard the mass of other staff, we are interested with TWICE only). One part of the track is a vertical loop with radius 12.0 m. At the bottom of the loop (point A) the car has speed 25.0 m/s and at the top of the loop (point B) it has speed 8.0 m's. As the car (with our TWICE idols) rolls from point A to point B, (a) how much work is done by friction? Using work energy theorem, (b) what is the gravitational potential energy? (c) what is the speed of the roller coaster car? (d) bonus: illustrate the problem. (5 %3D
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Transcribed Image Text:4. Gravitational Potential Energy. In TWICE
TV2, Dahyun (mass = 48.9 kg), Jeongyeon
(mass = 49.1 kg), and Chaeyoung (mass = 46
kg) were asked to ride an amusement park
roller coaster. They were testing this roller
coaster You are testing a new amusement park
roller coaster with a car of mass 120 kg
(disregard the mass of other staff, we are
interested with TWICE only). One part of the
track is a vertical loop with radius 12.0 m. At
the bottom of the loop (point A) the car has
speed 25.0 m/s and at the top of the loop (point B) it has speed 8.0 m/s. As the car (with our TWICE idols)
rolls from point A to point B, (a) how much work is done by friction? Using work energy theorem, (b)
what is the gravitational potential energy? (c) what is the speed of the roller coaster car? (d) bomus:
illustrate the problem. (
![4.
Given information:
The mass of the the staff:
m, = 48.9 kgm, = 49. 1 kgm, = 46 kg
Mass of the car (M) = 120 kg
The radius of the loop (1) = 12.0 m
The speed of the car at the bottom (point A) (V.) = 25.0 m/s
The speed at the top of the loop (point B) (V,) = 8.0 m/s
(a)
According to the work energy theorem the work done by friction is given as:
W, = (Mechanical energy at the bottom) – (Mechanical energy at the top)
W; = (mv,² + mg(0)) – (÷mv;² + mg(2R))
W, = (m, + m; + m, + M)[ (V,² – V,²) – 2gR]
Wy %3 (48.9 + 49. 1 +46 + 120) [을 (25-82)-2(9.8 x 12)]
W, = 11959. 2 kJ
(b)
The change in gravitational potential energy of the system is:
APE = mg2R
A PE = (48.9 + 49. 1+ 46+ 120) x 9. 8 x (2 × 12)
A PE = 10916. 4J
(c)
The speed of the car when it comes out the loop is given from work energy theorem as:
the final KE of the car = Initial KE – 2W;
글m? %3D글₩.? - 2W,
V,2 = V,2 - 4-
Im+m,+,+MI
11959.2
V, = V25? - 4
48.94-49.1+46+120
V = 21. 06 mls](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F01678e5f-48f5-4d64-aa90-28bc7b500125%2F2c4324f8-f0a4-4d3b-95cf-6dc280ceffc2%2F8a4hi0t_processed.jpeg&w=3840&q=75)
Transcribed Image Text:4.
Given information:
The mass of the the staff:
m, = 48.9 kgm, = 49. 1 kgm, = 46 kg
Mass of the car (M) = 120 kg
The radius of the loop (1) = 12.0 m
The speed of the car at the bottom (point A) (V.) = 25.0 m/s
The speed at the top of the loop (point B) (V,) = 8.0 m/s
(a)
According to the work energy theorem the work done by friction is given as:
W, = (Mechanical energy at the bottom) – (Mechanical energy at the top)
W; = (mv,² + mg(0)) – (÷mv;² + mg(2R))
W, = (m, + m; + m, + M)[ (V,² – V,²) – 2gR]
Wy %3 (48.9 + 49. 1 +46 + 120) [을 (25-82)-2(9.8 x 12)]
W, = 11959. 2 kJ
(b)
The change in gravitational potential energy of the system is:
APE = mg2R
A PE = (48.9 + 49. 1+ 46+ 120) x 9. 8 x (2 × 12)
A PE = 10916. 4J
(c)
The speed of the car when it comes out the loop is given from work energy theorem as:
the final KE of the car = Initial KE – 2W;
글m? %3D글₩.? - 2W,
V,2 = V,2 - 4-
Im+m,+,+MI
11959.2
V, = V25? - 4
48.94-49.1+46+120
V = 21. 06 mls
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