A 2 kg toy car sits at the highest point of a 13 m high hill. The car is gently pushed forward until it begins to roll down the slope. Assuming the car coasts freely, without any friction or air resistance, how much kinetic energy (KE) and potential energy (PE) will it have at each of the indicated points? Complete the diagram by placing the correct label in each bin. Use g = 10 m/s² for the acceleration due to gravity. The diagram is not drawn to scale. Answer Bank 4J 260 J 520 J 12 J OJ PE = 40 J 140 J 220 J 26 J 120 J KE = PE = KE = 13 m PE = KE = PE KE 6 m 2 m

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Chapter1: Units, Trigonometry. And Vectors
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**Description:**

The image illustrates a physics problem about energy conversion for a 2 kg toy car descending a 13 m high hill. The problem asks for the calculation of the kinetic energy (KE) and potential energy (PE) at various points as the car moves down the slope. It assumes no friction or air resistance, and the gravitational acceleration is given as \( g = 10 \, \text{m/s}^2 \).

**Diagram Details:**

- **Hill and Path:**
  - A toy car starts at the top of a 13 m high hill.
  - It moves down over two points: 2 m high and 6 m at the valley.

- **Calculation Points:**
  1. **At the Top (13 m):**
     - PE = 260 J
     - KE = 0 J

  2. **At 2 m Height:**
     - PE = 40 J
     - KE = 220 J

  3. **At 6 m Height:**
     - PE = 120 J
     - KE = 140 J

  4. **At the Bottom (0 m):**
     - PE = 0 J
     - KE = 260 J

**Answer Bank:**
- 0 J, 4 J, 12 J, 26 J, 40 J, 120 J, 140 J, 220 J, 260 J, 520 J

This problem visually represents the conservation of mechanical energy, illustrating how potential energy transforms into kinetic energy as the height decreases.
Transcribed Image Text:**Description:** The image illustrates a physics problem about energy conversion for a 2 kg toy car descending a 13 m high hill. The problem asks for the calculation of the kinetic energy (KE) and potential energy (PE) at various points as the car moves down the slope. It assumes no friction or air resistance, and the gravitational acceleration is given as \( g = 10 \, \text{m/s}^2 \). **Diagram Details:** - **Hill and Path:** - A toy car starts at the top of a 13 m high hill. - It moves down over two points: 2 m high and 6 m at the valley. - **Calculation Points:** 1. **At the Top (13 m):** - PE = 260 J - KE = 0 J 2. **At 2 m Height:** - PE = 40 J - KE = 220 J 3. **At 6 m Height:** - PE = 120 J - KE = 140 J 4. **At the Bottom (0 m):** - PE = 0 J - KE = 260 J **Answer Bank:** - 0 J, 4 J, 12 J, 26 J, 40 J, 120 J, 140 J, 220 J, 260 J, 520 J This problem visually represents the conservation of mechanical energy, illustrating how potential energy transforms into kinetic energy as the height decreases.
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