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A roller coaster car has a mass of 1300 kg is traveling at 17.8 m/s at position “A”. It coasts up to position “B”. What will be it’s velocity at position B, if B is vertically 14.0 meters higher and friction is ignored? 

User Pbrodka
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The velocity of the roller coaster car at position B, when it is vertically 14.0 meters higher and friction is ignored, is approximately 24.31 m/s.

To determine the velocity of the roller coaster car at position B, we can make use of the principle of conservation of mechanical energy. Since friction is ignored, the total mechanical energy of the system remains constant.

The total mechanical energy is the sum of the kinetic energy (KE) and the potential energy (PE) of the car.

At position A:

Initial kinetic energy (KEi) = 1/2 * mass * velocity^2

Given:

Mass of the car (m) = 1300 kg

Velocity at position A (vA) = 17.8 m/s

KEi = 1/2 * 1300 kg * (17.8 m/s)^2

At position B:

Final potential energy (PEf) = mass * gravitational acceleration * height

Given:

Height difference (Δh) = 14.0 meters

Gravitational acceleration (g) = 9.8 m/s^2 (approximately)

PEf = 1300 kg * 9.8 m/s^2 * 14.0 meters

According to the principle of conservation of mechanical energy, KEi + PEi = KEf + PEf.

Since we are ignoring friction, there is no change in kinetic energy, so KEi = KEf. Therefore, we can write:

PEi = PEf

Solving for the velocity at position B (vB), we can rearrange the equation as follows:

1/2 * mass * vA^2 + mass * g * 0 = 0 + mass * g * Δh

1/2 * 1300 kg * (17.8 m/s)^2 = 1300 kg * 9.8 m/s^2 * 14.0 meters + 1300 kg * g * Δh

vB = sqrt((2 * g * Δh) + vA^2)

Substituting the given values into the equation:

vB = sqrt((2 * 9.8 m/s^2 * 14.0 meters) + (17.8 m/s)^2)

vB = sqrt(274.4 + 316.84)

vB ≈ sqrt(591.24)

vB ≈ 24.31 m/s

Therefore, the velocity of the roller coaster car at position B, when it is vertically 14.0 meters higher and friction is ignored, is approximately 24.31 m/s.

User Demonsoul
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