Physical Science Worksheet on Conservation of Energy with diagrams and calculations for energy at different positions.

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):
$$
ME = KE + PE
$$

We assume no friction or air resistance, so mechanical energy is conserved throughout the motion.

---

Problem 1: Skater on a Ramp

Given:
- Mass $ m = 60 \, \text{kg} $
- At point ①: $ v = 8 \, \text{m/s} $, height $ h = 0 \, \text{m} $ (ground level)
- At point ②: height $ h = 1 \, \text{m} $
- At point ③: at the top of the ramp (height not labeled — we’ll calculate it from conservation)

#### Step 1: Calculate values at point ①

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

So:
- PE = 0 J
- KE = 1920 J
- ME = 1920 J
- v = 8 m/s
- h = 0 m

---

#### Step 2: Point ② (height = 1 m)

- $ h = 1 \, \text{m} $
- $ PE = mgh = 60 \times 9.8 \times 1 = 588 \, \text{J} $
- $ ME = 1920 \, \text{J} $ (conserved)
- So $ KE = ME - PE = 1920 - 588 = 1332 \, \text{J} $
- Use $ KE = \frac{1}{2}mv^2 $ to find $ v $:
$$
1332 = \frac{1}{2} \times 60 \times v^2 \Rightarrow 1332 = 30v^2 \Rightarrow v^2 = \frac{1332}{30} = 44.4
\Rightarrow v = \sqrt{44.4} \approx 6.67 \, \text{m/s}
$$

So:
- PE = 588 J
- KE = 1332 J
- ME = 1920 J
- v ≈ 6.67 m/s
- h = 1 m

---

#### Step 3: Point ③ (top of ramp)

At point ③, the skater has maximum height, so KE = 0 (momentarily at rest), and all energy is potential.

- $ ME = 1920 \, \text{J} $
- $ KE = 0 $
- So $ PE = 1920 \, \text{J} $
- Use $ PE = mgh $ to find $ h $:
$$
1920 = 60 \times 9.8 \times h \Rightarrow h = \frac{1920}{588} \approx 3.26 \, \text{m}
$$

So:
- PE = 1920 J
- KE = 0 J
- ME = 1920 J
- v = 0 m/s
- h ≈ 3.26 m

---

✔ 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 | ~6.67 | 1 |
| ③ | 1920 | 0 | 1920 | 0 | ~3.26 |

---

Problem 2: Ball on Track

Given:
- $ m = 50 \, \text{kg} $
- At point ①: height = 4 m, velocity = 0 (assumed at start, at rest)
- At point ②: lowest point, height = 0
- At point ③: loop, height = 3 m
- At point ④: final hill, $ v = 6 \, \text{m/s} $, height = ?

We'll compute values at each point using conservation of mechanical energy.

---

#### Step 1: Point ① (start)

- $ h = 4 \, \text{m} $
- $ v = 0 $ → $ KE = 0 $
- $ PE = mgh = 50 \times 9.8 \times 4 = 1960 \, \text{J} $
- $ ME = 0 + 1960 = 1960 \, \text{J} $

So:
- PE = 1960 J
- KE = 0 J
- ME = 1960 J
- v = 0 m/s
- h = 4 m

---

#### Step 2: Point ② (bottom of track, h = 0)

- $ h = 0 $ → $ PE = 0 $
- $ ME = 1960 \, \text{J} $
- So $ KE = 1960 \, \text{J} $
- $ KE = \frac{1}{2}mv^2 \Rightarrow 1960 = \frac{1}{2} \times 50 \times v^2 = 25v^2 $
$$
v^2 = \frac{1960}{25} = 78.4 \Rightarrow v = \sqrt{78.4} \approx 8.85 \, \text{m/s}
$$

So:
- PE = 0 J
- KE = 1960 J
- ME = 1960 J
- v ≈ 8.85 m/s
- h = 0 m

---

#### Step 3: Point ③ (loop, h = 3 m)

- $ h = 3 \, \text{m} $
- $ PE = 50 \times 9.8 \times 3 = 1470 \, \text{J} $
- $ ME = 1960 \, \text{J} $
- $ KE = 1960 - 1470 = 490 \, \text{J} $
- $ KE = \frac{1}{2}mv^2 \Rightarrow 490 = \frac{1}{2} \times 50 \times v^2 = 25v^2 $
$$
v^2 = \frac{490}{25} = 19.6 \Rightarrow v = \sqrt{19.6} \approx 4.43 \, \text{m/s}
$$

So:
- PE = 1470 J
- KE = 490 J
- ME = 1960 J
- v ≈ 4.43 m/s
- h = 3 m

---

#### Step 4: Point ④ (final hill, v = 6 m/s)

- Given: $ v = 6 \, \text{m/s} $
- $ KE = \frac{1}{2} \times 50 \times (6)^2 = 25 \times 36 = 900 \, \text{J} $
- $ ME = 1960 \, \text{J} $
- So $ PE = ME - KE = 1960 - 900 = 1060 \, \text{J} $
- $ PE = mgh \Rightarrow 1060 = 50 \times 9.8 \times h $
$$
h = \frac{1060}{490} \approx 2.16 \, \text{m}
$$

So:
- PE = 1060 J
- KE = 900 J
- ME = 1960 J
- v = 6 m/s
- h ≈ 2.16 m

---

✔ Summary for Problem 2:

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

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Final Notes:

- In both problems, mechanical energy is conserved.
- The key idea is that as height increases, potential energy increases and kinetic energy decreases (and vice versa).
- We used $ g = 9.8 \, \text{m/s}^2 $, but sometimes problems use $ g = 10 \, \text{m/s}^2 $. If your teacher uses $ g = 10 $, just recalculate with that value.

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