Question

A roller coaster car has a mass of 250 kg. the car is sitting on top of a 25 m hill. the gravitational potential energy of the car at the highest point of the track is 61,250 j. ​ kinetic energy can be modeled by the formula , where m is the mass of an object and v is the velocity of an object. ​ what is the maximum velocity of the roller coaster car at the lowest point of the track? round the answer to the nearest whole number. a

Answer 1

The maximum velocity of the roller coaster car at the lowest point of the track is 22 m/s.

Step-by-step explanation:

The maximum velocity of the roller coaster car at the lowest point of the track can be calculated using the principle of conservation of energy. The roller coaster car initially has gravitational potential energy at the top of the 25 m hill. This potential energy is converted into kinetic energy at the lowest point of the track. According to the conservation of energy, the loss of gravitational potential energy is equal to the gain in kinetic energy. So, we can set the initial potential energy equal to the final kinetic energy:

APEg = AKE

-mg |h| = \dfrac{1}{2}mv^2

We know that the mass (m) of the car is 250 kg, the gravitational potential energy (APEg) is 61,250 J, and the height (h) is 25 m. Rearranging the equation and substituting the values, we get:

v^2 = \dfrac{-2 \times 61,250}{250}

v^2 = -490

The velocity cannot be negative, so we take the square root of the absolute value to get:

v = \sqrt{490}

Rounding the answer to the nearest whole number, the maximum velocity of the roller coaster car at the lowest point of the track is 22 m/s.

Answer 2

To find the maximum velocity of the roller coaster car at the lowest point, the gravitational potential energy at the top is equated with the kinetic energy at the bottom, and by solving the equations, the maximum velocity is approximately 22 m/s.

Step-by-step explanation:

The student's question is about finding the maximum velocity of a roller coaster car at the lowest point of the track, given that its mass is 250 kg and the gravitational potential energy (GPE) at the top is 61,250 J. According to the conservation of energy, the GPE at the top of the hill will convert into kinetic energy (KE) at the bottom of the hill if we neglect energy losses due to friction. The formula for KE is KE = ½ mv², where m is mass and v is velocity. We are given that GPE = mgh, where g is the acceleration due to gravity (9.8 m/s²) and h is the height (25 m). Because the total mechanical energy is conserved, the GPE at the top will be equal to the KE at the bottom; therefore, we can set them equal to each other and solve for v:

GPE = KE
mg(h) = ½ mv²
61,250 J = ½ × 250 kg × v²

By simplifying this equation and solving for v, we find that:

61,250 J = 125 kg × v²
490 = v²
v ≈ √490 ≈ 22.1 m/s

The maximum velocity of the roller coaster car at the lowest point of the track is approximately 22 m/s, which rounds to 22 m/s as the nearest whole number.