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A 440 kg roller coaster car is going [tex]26 \, \text{m/s}[/tex] when it reaches the lowest point on the track. If the car started from rest at the top of a hill, how much higher was that point on the track than the lowest point? (Use [tex]g=9.80 \, \text{m/s}^2[/tex], and ignore friction.)

A. 17 m
B. 23 m
C. 34 m
D. 69 m


Sagot :

To determine the height difference between the starting point and the lowest point on a track for a roller coaster car, we can use the principle of conservation of mechanical energy. This principle states that the total mechanical energy (potential energy + kinetic energy) in a closed system remains constant if only conservative forces are acting (ignoring friction).

Given:
- Mass of the roller coaster car, [tex]\( m = 440 \)[/tex] kg
- Final velocity at the lowest point, [tex]\( v = 26 \)[/tex] m/s
- Acceleration due to gravity, [tex]\( g = 9.8 \)[/tex] m/s[tex]\(^2\)[/tex]

Step-by-Step Solution:

1. Determine the kinetic energy at the lowest point:

The kinetic energy ([tex]\( KE \)[/tex]) at the lowest point is given by the formula:
[tex]\[ KE = \frac{1}{2}mv^2 \][/tex]
Plugging in the values:
[tex]\[ KE = \frac{1}{2} \times 440 \, \text{kg} \times (26 \, \text{m/s})^2 \][/tex]
Solving this:
[tex]\[ KE = 0.5 \times 440 \times 676 \][/tex]
[tex]\[ KE = 0.5 \times 297440 \][/tex]
[tex]\[ KE = 148720 \, \text{J} \][/tex]

2. Convert the kinetic energy to potential energy at the top:

At the top of the hill, the roller coaster car starts from rest, so its kinetic energy is zero and all the mechanical energy is in the form of potential energy ([tex]\( PE \)[/tex]). According to the conservation of energy:
[tex]\[ PE_{top} = KE_{bottom} \][/tex]
The potential energy at a height [tex]\( h \)[/tex] is given by:
[tex]\[ PE = mgh \][/tex]
Setting this equal to the kinetic energy calculated:
[tex]\[ mgh = 148720 \, \text{J} \][/tex]

3. Solve for the height [tex]\( h \)[/tex]:

[tex]\[ h = \frac{KE}{mg} \][/tex]
Substituting the given values:
[tex]\[ h = \frac{148720 \, \text{J}}{440 \, \text{kg} \times 9.8 \, \text{m/s}^2} \][/tex]
[tex]\[ h = \frac{148720}{4312} \][/tex]
[tex]\[ h \approx 34.49 \, \text{m} \][/tex]

Therefore, the height difference between the starting point at the top of the hill and the lowest point on the track is approximately 34 meters, making the correct answer:

34 m
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