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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} m v^2[/tex].

A kiddie roller coaster car has a mass of 100 kilograms. At the top of a hill, it is 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 double its kinetic energy at the top. The car has joules of kinetic energy at the bottom of the hill.



Answer :

Let's solve the problem step by step:

1. Understanding the Problem:
- Mass of the roller coaster car, [tex]\( m \)[/tex] = 100 kg.
- Speed at the top of the hill, [tex]\( v_{\text{top}} \)[/tex] = 3 m/s.
- Speed at the bottom of the hill, [tex]\( v_{\text{bottom}} \)[/tex] = 2 [tex]\( v_{\text{top}} \)[/tex] = 2 3 m/s = 6 m/s.

2. Kinetic Energy Formula:
- The formula for kinetic energy (KE) is [tex]\( KE = \frac{1}{2} m v^2 \)[/tex].

3. Calculating Kinetic Energy at the Top:
- [tex]\( KE_{\text{top}} = \frac{1}{2} m v_{\text{top}}^2 \)[/tex].
- Substituting the given values:
[tex]\[ KE_{\text{top}} = \frac{1}{2} \times 100 \times 3^2 = \frac{1}{2} \times 100 \times 9 = 50 \times 9 = 450 \text{ J} \][/tex]

4. Calculating Kinetic Energy at the Bottom:
- [tex]\( KE_{\text{bottom}} = \frac{1}{2} m v_{\text{bottom}}^2 \)[/tex].
- Substituting the given values:
[tex]\[ KE_{\text{bottom}} = \frac{1}{2} \times 100 \times 6^2 = \frac{1}{2} \times 100 \times 36 = 50 \times 36 = 1800 \text{ J} \][/tex]

So, the kinetic energy of the roller coaster car at the top of the hill is 450 J, and at the bottom of the hill, it is 1800 J. The car's kinetic energy at the bottom is 1800 joules.

Therefore, the correct answers are:
- The car's kinetic energy at the bottom is 1800 joules.
- The car's kinetic energy at the top is 450 joules.