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9. Lee Ben Fardest (esteemed American ski jumper) has a mass of [tex]$59.6 \, \text{kg}[tex]$[/tex]. He is moving with a speed of [tex]$[/tex]23.4 \, \text{m/s}[tex]$[/tex] at a height of [tex]$[/tex]44.6 \, \text{meters}$[/tex] above the ground. Determine the total mechanical energy of Lee Ben Fardest.

Sagot :

Certainly! To determine the total mechanical energy of Lee Ben Fardest, we need to calculate both his kinetic energy (KE) and his potential energy (PE). The total mechanical energy (TME) is the sum of these two forms of energy.

### 1. Kinetic Energy (KE)
The formula for kinetic energy is:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
where:
- \( m \) is the mass,
- \( v \) is the speed of the object.

Given:
- \( m = 59.6 \, \text{kg} \)
- \( v = 23.4 \, \text{m/s} \)

Substitute the given values into the kinetic energy formula:
[tex]\[ KE = \frac{1}{2} \times 59.6 \, \text{kg} \times (23.4 \, \text{m/s})^2 \][/tex]
[tex]\[ KE = \frac{1}{2} \times 59.6 \times 547.56 \][/tex]
[tex]\[ KE = \frac{1}{2} \times 32646.336 \][/tex]
[tex]\[ KE = 16317.288 \, \text{J} \][/tex]

### 2. Potential Energy (PE)
The formula for potential energy is:
[tex]\[ PE = m g h \][/tex]
where:
- \( m \) is the mass,
- \( g \) is the acceleration due to gravity (approximated as \( 9.8 \, \text{m/s}^2 \)),
- \( h \) is the height above the ground.

Given:
- \( m = 59.6 \, \text{kg} \)
- \( g = 9.8 \, \text{m/s}^2 \)
- \( h = 44.6 \, \text{m} \)

Substitute the given values into the potential energy formula:
[tex]\[ PE = 59.6 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 44.6 \, \text{m} \][/tex]
[tex]\[ PE = 59.6 \times 437.08 \][/tex]
[tex]\[ PE = 26049.968 \, \text{J} \][/tex]

### 3. Total Mechanical Energy (TME)
The total mechanical energy is the sum of the kinetic energy and the potential energy:
[tex]\[ TME = KE + PE \][/tex]
[tex]\[ TME = 16317.288 \, \text{J} + 26049.968 \, \text{J} \][/tex]
[tex]\[ TME = 42367.256 \, \text{J} \][/tex]

Therefore, the total mechanical energy of Lee Ben Fardest is:
[tex]\[ 42367.256 \, \text{J} \][/tex]