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What is the kinetic energy of the roller coaster at the top and bottom of the hill? Use [tex]$KE = \frac{1}{2}mv^2$[/tex].

A kiddie roller coaster car has a mass of 100 kilograms. At the top of a hill, it's moving at a speed of 3 meters/second. After reaching the bottom of the hill, its speed doubles. The car's kinetic energy at the bottom is [tex]\square[/tex] its kinetic energy at the top. The car has [tex]\square[/tex] joules of kinetic energy at the bottom of the hill.



Answer :

Let's go through the solution step-by-step:

1. Given Data:
- Mass of the roller coaster car, \( m = 100 \) kilograms.
- Speed at the top of the hill, \( v_{\text{top}} = 3 \) meters/second.
- Speed at the bottom of the hill, \( v_{\text{bottom}} = 2 \times v_{\text{top}} = 2 \times 3 = 6 \) meters/second.

2. Kinetic Energy Calculation:
The formula for kinetic energy is given by:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]

3. Kinetic Energy at the Top:
Substitute the values \( m = 100 \) kg and \( v_{\text{top}} = 3 \) m/s into the formula:
[tex]\[ KE_{\text{top}} = \frac{1}{2} \times 100 \times (3^2) = \frac{1}{2} \times 100 \times 9 = 450 \text{ joules} \][/tex]

4. Kinetic Energy at the Bottom:
Substitute the values \( m = 100 \) kg and \( v_{\text{bottom}} = 6 \) m/s into the formula:
[tex]\[ KE_{\text{bottom}} = \frac{1}{2} \times 100 \times (6^2) = \frac{1}{2} \times 100 \times 36 = 1800 \text{ joules} \][/tex]

5. Ratio of Kinetic Energies:
To find how many times greater the kinetic energy at the bottom is compared to the kinetic energy at the top:
[tex]\[ \text{Ratio} = \frac{KE_{\text{bottom}}}{KE_{\text{top}}} = \frac{1800}{450} = 4 \][/tex]

So, based on the calculations:

- The car's kinetic energy at the bottom is 4 times its kinetic energy at the top.
- The car has 1800 joules of kinetic energy at the bottom of the hill.

Let's fill the blanks with the correct answers.

"The car's kinetic energy at the bottom is 4 times its kinetic energy at the top. The car has 1800 joules of kinetic energy at the bottom of the hill."