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

Problem Analysis:

The worksheet involves calculating potential energy (PE), kinetic energy (KE), mechanical energy (ME), velocity (v), and height (h) at various points for two scenarios:
1. A skater on a ramp.
2. A ball in motion.

We will solve each part step by step using the given formulas:
- Potential Energy (PE): \( PE = mgh \)
- Kinetic Energy (KE): \( KE = \frac{1}{2}mv^2 \)
- Mechanical Energy (ME): \( ME = PE + KE \)

Part 1: Skater on a Ramp

#### Given:
- Mass of the skater, \( m = 60 \, \text{kg} \)
- Initial velocity at point ①, \( v = 8 \, \text{m/s} \)
- Height at point ②, \( h = 1 \, \text{m} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

#### Step-by-Step Solution:

##### Point ① (Ground Level):
- Height (\( h \)): \( h = 0 \, \text{m} \)
- Potential Energy (PE):
\[
PE = mgh = 60 \times 9.8 \times 0 = 0 \, \text{J}
\]
- Kinetic Energy (KE):
\[
KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 60 \times (8)^2 = 30 \times 64 = 1920 \, \text{J}
\]
- Mechanical Energy (ME):
\[
ME = PE + KE = 0 + 1920 = 1920 \, \text{J}
\]

##### Point ② (1 m Above Ground):
- Height (\( h \)): \( h = 1 \, \text{m} \)
- Potential Energy (PE):
\[
PE = mgh = 60 \times 9.8 \times 1 = 588 \, \text{J}
\]
- Mechanical Energy (ME): Since mechanical energy is conserved, \( ME = 1920 \, \text{J} \).
- Kinetic Energy (KE):
\[
KE = ME - PE = 1920 - 588 = 1332 \, \text{J}
\]
- Velocity (\( v \)):
\[
KE = \frac{1}{2}mv^2 \implies 1332 = \frac{1}{2} \times 60 \times v^2
\]
\[
1332 = 30v^2 \implies v^2 = \frac{1332}{30} = 44.4 \implies v = \sqrt{44.4} \approx 6.66 \, \text{m/s}
\]

##### Point ③ (Highest Point):
- At the highest point, the skater's velocity is zero (\( v = 0 \)).
- Kinetic Energy (KE):
\[
KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 60 \times 0^2 = 0 \, \text{J}
\]
- Mechanical Energy (ME): Since mechanical energy is conserved, \( ME = 1920 \, \text{J} \).
- Potential Energy (PE):
\[
PE = ME - KE = 1920 - 0 = 1920 \, \text{J}
\]
- Height (\( h \)):
\[
PE = mgh \implies 1920 = 60 \times 9.8 \times h
\]
\[
1920 = 588h \implies h = \frac{1920}{588} \approx 3.27 \, \text{m}
\]

#### Summary for Skater:
| Point | \( h \) (m) | \( PE \) (J) | \( KE \) (J) | \( ME \) (J) | \( v \) (m/s) |
|-------|-------------|--------------|--------------|--------------|---------------|
| ① | 0 | 0 | 1920 | 1920 | 8 |
| ② | 1 | 588 | 1332 | 1920 | 6.66 |
| ③ | 3.27 | 1920 | 0 | 1920 | 0 |

---

Part 2: Ball in Motion

#### Given:
- Mass of the ball, \( m = 59 \, \text{kg} \)
- Initial height at point ①, \( h = 4 \, \text{m} \)
- Velocity at point ④, \( v = 6 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

#### Step-by-Step Solution:

##### Point ① (Initial Position):
- Height (\( h \)): \( h = 4 \, \text{m} \)
- Potential Energy (PE):
\[
PE = mgh = 59 \times 9.8 \times 4 = 2312.8 \, \text{J}
\]
- Kinetic Energy (KE): At the start, the ball is at rest (\( v = 0 \)).
\[
KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 59 \times 0^2 = 0 \, \text{J}
\]
- Mechanical Energy (ME):
\[
ME = PE + KE = 2312.8 + 0 = 2312.8 \, \text{J}
\]

##### Point ② (Mid-Air):
- Height (\( h \)): \( h = 3 \, \text{m} \)
- Potential Energy (PE):
\[
PE = mgh = 59 \times 9.8 \times 3 = 1734.6 \, \text{J}
\]
- Mechanical Energy (ME): Since mechanical energy is conserved, \( ME = 2312.8 \, \text{J} \).
- Kinetic Energy (KE):
\[
KE = ME - PE = 2312.8 - 1734.6 = 578.2 \, \text{J}
\]
- Velocity (\( v \)):
\[
KE = \frac{1}{2}mv^2 \implies 578.2 = \frac{1}{2} \times 59 \times v^2
\]
\[
578.2 = 29.5v^2 \implies v^2 = \frac{578.2}{29.5} \approx 19.6 \implies v = \sqrt{19.6} \approx 4.43 \, \text{m/s}
\]

##### Point ③ (Ground Level):
- Height (\( h \)): \( h = 0 \, \text{m} \)
- Potential Energy (PE):
\[
PE = mgh = 59 \times 9.8 \times 0 = 0 \, \text{J}
\]
- Mechanical Energy (ME): Since mechanical energy is conserved, \( ME = 2312.8 \, \text{J} \).
- Kinetic Energy (KE):
\[
KE = ME - PE = 2312.8 - 0 = 2312.8 \, \text{J}
\]
- Velocity (\( v \)):
\[
KE = \frac{1}{2}mv^2 \implies 2312.8 = \frac{1}{2} \times 59 \times v^2
\]
\[
2312.8 = 29.5v^2 \implies v^2 = \frac{2312.8}{29.5} \approx 78.4 \implies v = \sqrt{78.4} \approx 8.85 \, \text{m/s}
\]

##### Point ④ (Rising Again):
- Velocity (\( v \)): Given as \( v = 6 \, \text{m/s} \)
- Kinetic Energy (KE):
\[
KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 59 \times 6^2 = 29.5 \times 36 = 1062 \, \text{J}
\]
- Mechanical Energy (ME): Since mechanical energy is conserved, \( ME = 2312.8 \, \text{J} \).
- Potential Energy (PE):
\[
PE = ME - KE = 2312.8 - 1062 = 1250.8 \, \text{J}
\]
- Height (\( h \)):
\[
PE = mgh \implies 1250.8 = 59 \times 9.8 \times h
\]
\[
1250.8 = 588h \implies h = \frac{1250.8}{588} \approx 2.13 \, \text{m}
\]

#### Summary for Ball:
| Point | \( h \) (m) | \( PE \) (J) | \( KE \) (J) | \( ME \) (J) | \( v \) (m/s) |
|-------|-------------|--------------|--------------|--------------|---------------|
| ① | 4 | 2312.8 | 0 | 2312.8 | 0 |
| ② | 3 | 1734.6 | 578.2 | 2312.8 | 4.43 |
| ③ | 0 | 0 | 2312.8 | 2312.8 | 8.85 |
| ④ | 2.13 | 1250.8 | 1062 | 2312.8 | 6 |

---

Final Answers:

1. Skater:
\[
\boxed{
\begin{array}{c|c|c|c|c|c}
\text{Point} & h \, (\text{m}) & PE \, (\text{J}) & KE \, (\text{J}) & ME \, (\text{J}) & v \, (\text{m/s}) \\
\hline
① & 0 & 0 & 1920 & 1920 & 8 \\
② & 1 & 588 & 1332 & 1920 & 6.66 \\
③ & 3.27 & 1920 & 0 & 1920 & 0 \\
\end{array}
}
\]

2. Ball:
\[
\boxed{
\begin{array}{c|c|c|c|c|c}
\text{Point} & h \, (\text{m}) & PE \, (\text{J}) & KE \, (\text{J}) & ME \, (\text{J}) & v \, (\text{m/s}) \\
\hline
① & 4 & 2312.8 & 0 & 2312.8 & 0 \\
② & 3 & 1734.6 & 578.2 & 2312.8 & 4.43 \\
③ & 0 & 0 & 2312.8 & 2312.8 & 8.85 \\
④ & 2.13 & 1250.8 & 1062 & 2312.8 & 6 \\
\end{array}
}
\]