Conservation of Energy | PDF | Kinetic Energy | Potential Energy - Free Printable

Let's solve the Conservation of Energy worksheet step by step. We'll use the following formulas:

- Kinetic Energy (KE): $ KE = \frac{1}{2}mv^2 $
- Gravitational Potential Energy (GPE or PE): $ PE = mgh $
- Mechanical Energy (ME): $ ME = KE + PE $
- Velocity (v): Use energy conservation to find unknown velocities.
- Height (h): Use energy conservation to find unknown heights.

We assume:
- Acceleration due to gravity: $ g = 9.8 \, \text{m/s}^2 $
- No friction or air resistance (ideal case), so mechanical energy is conserved.

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Problem 1: Skateboarder on a Ramp

Given:
- Mass $ m = 60 \, \text{kg} $
- At point ①: $ v = 8 \, \text{m/s} $, height $ h = 0 \, \text{m} $

#### Point ① (Bottom, starting point):
- $ h = 0 $ → $ PE = mgh = 60 \times 9.8 \times 0 = 0 \, \text{J} $
- $ KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 60 \times (8)^2 = 30 \times 64 = 1920 \, \text{J} $
- $ ME = KE + PE = 1920 + 0 = 1920 \, \text{J} $
- $ v = 8 \, \text{m/s} $ (given)
- $ h = 0 \, \text{m} $ (given)

✔ So at point ①:
- $ PE = 0 \, \text{J} $
- $ KE = 1920 \, \text{J} $
- $ ME = 1920 \, \text{J} $
- $ v = 8 \, \text{m/s} $
- $ h = 0 \, \text{m} $

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#### Point ② (Midway up ramp, height = 1 m):
- $ h = 1 \, \text{m} $
- $ PE = mgh = 60 \times 9.8 \times 1 = 588 \, \text{J} $
- Since mechanical energy is conserved: $ ME = 1920 \, \text{J} $
- So $ KE = ME - PE = 1920 - 588 = 1332 \, \text{J} $
- Now find velocity:
$ KE = \frac{1}{2}mv^2 \Rightarrow v^2 = \frac{2 \times KE}{m} = \frac{2 \times 1332}{60} = \frac{2664}{60} = 44.4 $
$ v = \sqrt{44.4} \approx 6.66 \, \text{m/s} $

✔ So at point ②:
- $ PE = 588 \, \text{J} $
- $ KE = 1332 \, \text{J} $
- $ ME = 1920 \, \text{J} $
- $ v \approx 6.66 \, \text{m/s} $
- $ h = 1 \, \text{m} $

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#### Point ③ (Top of ramp, highest point):
- Let’s suppose the skater reaches the top with zero velocity (momentarily stops).
- But we need to calculate based on conservation.

Wait — we don’t know the height at ③, but we can find it using total ME.

At point ③:
- If the skater reaches maximum height, then $ KE = 0 $ at that instant.
- So all energy is potential: $ PE = ME = 1920 \, \text{J} $
- $ PE = mgh \Rightarrow h = \frac{PE}{mg} = \frac{1920}{60 \times 9.8} = \frac{1920}{588} \approx 3.26 \, \text{m} $

So:
- $ PE = 1920 \, \text{J} $
- $ KE = 0 \, \text{J} $
- $ ME = 1920 \, \text{J} $
- $ v = 0 \, \text{m/s} $
- $ h \approx 3.26 \, \text{m} $

But wait — in the diagram, point ③ is higher than point ②, and the ramp goes up from ② to ③. But we were not given height at ③.

However, since mechanical energy is conserved and we know the initial kinetic energy is 1920 J, the maximum possible height occurs when KE = 0.

So yes, at point ③ (top), $ KE = 0 $, $ PE = 1920 \, \text{J} $, $ h = \frac{1920}{60 \times 9.8} \approx 3.26 \, \text{m} $

✔ So at point ③:
- $ PE = 1920 \, \text{J} $
- $ KE = 0 \, \text{J} $
- $ ME = 1920 \, \text{J} $
- $ v = 0 \, \text{m/s} $
- $ h \approx 3.26 \, \text{m} $

> Note: The diagram may imply point ③ is just above point ②, but unless otherwise specified, we assume it's the highest point where speed becomes zero.

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✔ Summary for Problem 1:

| Location | PE (J) | KE (J) | ME (J) | v (m/s) | h (m) |
|--------|--------|--------|--------|--------|-------|
| ① | 0 | 1920 | 1920 | 8 | 0 |
| ② | 588 | 1332 | 1920 | ~6.66 | 1 |
| ③ | 1920 | 0 | 1920 | 0 | ~3.26 |

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Problem 2: Ball on Track

Given:
- Mass $ m = 50 \, \text{kg} $
- Starts at point ①: $ h = 4 \, \text{m} $, $ v = 0 $ (assumed at rest at top)
- Point ④: $ v = 6 \, \text{m/s} $, $ h = ? $

We’ll go through each point.

#### Point ① (Top of ramp):
- $ h = 4 \, \text{m} $
- $ PE = mgh = 50 \times 9.8 \times 4 = 1960 \, \text{J} $
- $ KE = \frac{1}{2}mv^2 = 0 $ (since $ v = 0 $)
- $ ME = KE + PE = 0 + 1960 = 1960 \, \text{J} $
- $ v = 0 \, \text{m/s} $
- $ h = 4 \, \text{m} $

✔ So at ①:
- $ PE = 1960 \, \text{J} $
- $ KE = 0 \, \text{J} $
- $ ME = 1960 \, \text{J} $
- $ v = 0 \, \text{m/s} $
- $ h = 4 \, \text{m} $

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#### Point ② (Top of loop, height = 3 m):
- $ h = 3 \, \text{m} $
- $ PE = mgh = 50 \times 9.8 \times 3 = 1470 \, \text{J} $
- $ ME = 1960 \, \text{J} $ (conserved)
- $ KE = ME - PE = 1960 - 1470 = 490 \, \text{J} $
- $ v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 490}{50}} = \sqrt{\frac{980}{50}} = \sqrt{19.6} \approx 4.43 \, \text{m/s} $

✔ So at ②:
- $ PE = 1470 \, \text{J} $
- $ KE = 490 \, \text{J} $
- $ ME = 1960 \, \text{J} $
- $ v \approx 4.43 \, \text{m/s} $
- $ h = 3 \, \text{m} $

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#### Point ③ (Bottom of loop, height = 0 m):
- $ h = 0 \, \text{m} $
- $ PE = 0 $
- $ ME = 1960 \, \text{J} $
- $ KE = 1960 \, \text{J} $
- $ v = \sqrt{\frac{2 \times 1960}{50}} = \sqrt{\frac{3920}{50}} = \sqrt{78.4} \approx 8.85 \, \text{m/s} $

✔ So at ③:
- $ PE = 0 \, \text{J} $
- $ KE = 1960 \, \text{J} $
- $ ME = 1960 \, \text{J} $
- $ v \approx 8.85 \, \text{m/s} $
- $ h = 0 \, \text{m} $

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#### Point ④ (Top of hill, $ v = 6 \, \text{m/s} $, $ h = ? $):
- $ v = 6 \, \text{m/s} $
- $ KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 50 \times 36 = 25 \times 36 = 900 \, \text{J} $
- $ ME = 1960 \, \text{J} $
- $ PE = ME - KE = 1960 - 900 = 1060 \, \text{J} $
- $ PE = mgh \Rightarrow h = \frac{PE}{mg} = \frac{1060}{50 \times 9.8} = \frac{1060}{490} \approx 2.16 \, \text{m} $

✔ So at ④:
- $ PE = 1060 \, \text{J} $
- $ KE = 900 \, \text{J} $
- $ ME = 1960 \, \text{J} $
- $ v = 6 \, \text{m/s} $
- $ h \approx 2.16 \, \text{m} $

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✔ Summary for Problem 2:

| Location | PE (J) | KE (J) | ME (J) | v (m/s) | h (m) |
|--------|--------|--------|--------|--------|-------|
| ① | 1960 | 0 | 1960 | 0 | 4 |
| ② | 1470 | 490 | 1960 | ~4.43 | 3 |
| ③ | 0 | 1960 | 1960 | ~8.85 | 0 |
| ④ | 1060 | 900 | 1960 | 6 | ~2.16 |

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✔ Final Answers

#### Problem 1: Skater

| Point | PE (J) | KE (J) | ME (J) | v (m/s) | h (m) |
|-------|--------|--------|--------|--------|-------|
| ① | 0 | 1920 | 1920 | 8 | 0 |
| ② | 588 | 1332 | 1920 | 6.66 | 1 |
| ③ | 1920 | 0 | 1920 | 0 | 3.26 |

#### Problem 2: Ball

| Point | PE (J) | KE (J) | ME (J) | v (m/s) | h (m) |
|-------|--------|--------|--------|--------|-------|
| ① | 1960 | 0 | 1960 | 0 | 4 |
| ② | 1470 | 490 | 1960 | 4.43 | 3 |
| ③ | 0 | 1960 | 1960 | 8.85 | 0 |
| ④ | 1060 | 900 | 1960 | 6 | 2.16 |

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🔍 Key Concept:

Conservation of Mechanical Energy: In the absence of non-conservative forces (like friction), the total mechanical energy (KE + PE) remains constant.

We used this principle to find missing values at each point.

Let me know if you'd like this formatted as a printable answer sheet!