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Dafna1 [17]
3 years ago
6

a roller coaster begins at the top of a hill if it acelerates at the rate of 2 m/s2 and has a mass of 2000 kg what net force is

acting on it plzzz someone help im over here crying bc i dont know
Physics
1 answer:
AlladinOne [14]3 years ago
7 0

Answer:

\boxed {\boxed {\sf 4000 \ Newtons }}

Explanation:

Force can be found by multiplying the mass by the acceleration.

F=m*a

The mass of the roller coaster is 2000 kilograms and the acceleration is 2 meters per second squared.

m= 2000 \ kg \\a= 2 \ m/s^2

Substitute the values into the formula.

F= 2000 \ kg * 2 \ m/s^2

Multiply.

F= 4000  \ kg*m/s^2

  • 1 kg*m/s² is equal to 1 N
  • Therefore our answer of 4000 kg*m/s² is equal to 4000 Newtons

F= 4000 \ N

The net force acting on the roller coaster is <u>4000 Newtons.</u>

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a. i. 3.43 m/s ii. 2.8 m/s

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a. Find translational speed of each cylinder upon reaching the bottom

The potential energy change of each mass = total kinetic energy gain = translational kinetic energy + rotational kinetic energy

So, mgh = 1/2mv² + 1/2Iω² where m = mass of object = 2.0 kg, g =acceleration due to gravity = 9.8 m/s², h = height of incline = 1.2 m, v = translational velocity of object, I = moment of inertia of object and ω = angular speed = v/r where r = radius of object.

i. translational speed of thin-walled cylinder upon reaching the bottom

So, For the thin-walled cylinder, I = mr², we find its translational velocity, v

So, mgh = 1/2mv² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²)(v/r)²  

mgh = 1/2mv² + 1/2mv²

mgh = mv²

v² = gh

v = √gh

v = √(9.8 m/s² × 1.2 m)

v = √(11.76 m²/s²)

v = 3.43 m/s

ii. translational speed of solid cylinder upon reaching the bottom

So, For the solid cylinder, I = mr²/2, we find its translational velocity, v'

So, mgh = 1/2mv'² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²/2)(v'/r)²  

mgh = 1/2mv'² + mv'²

mgh = 3mv'²/2

v'² = 2gh/3

v' = √(2gh/3)

v' = √(2 × 9.8 m/s² × 1.2 m/3)

v' = √(23.52 m²/s²/3)

v' = √(7.84 m²/s²)

v' = 2.8 m/s

b. Determine which cylinder has the greatest translational speed upon reaching the bottom.

Since v = 3.43 m/s > v'= 2.8 m/s,

the thin-walled cylinder has the greatest translational speed upon reaching the bottom.

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