To determine the height of the watermelon above the ground when its kinetic energy is 4.61 kJ, we can use the principle of conservation of energy.
The total mechanical energy consists of potential energy (PE) and kinetic energy (KE):
\[ \text{Total Energy} = \text{PE} + \text{KE} \]
Initially, the watermelon has potential energy when it is at a height \( h_i = 5.00 \) m. The potential energy can be calculated using the formula:
\[ \text{PE}_i = mgh_i \]
where:
- \( m = 118 \) kg (mass of the watermelon),
- \( g = 9.81 \) m/s² (acceleration due to gravity),
- \( h_i = 5.00 \) m.
Now, substituting the values in to find the initial potential energy:
\[ \text{PE}_i = 118 , \text{kg} \times 9.81 , \text{m/s}^2 \times 5.00 , \text{m} \]
\[ \text{PE}_i = 118 \times 9.81 \times 5.00 \approx 5796.90 , \text{J} \]
Now, converting this to kJ:
\[ \text{PE}_i \approx 5.80 , \text{kJ} \]
When the watermelon slides down the ramp and has a kinetic energy of \( KE = 4.61 \) kJ, we can use the conservation of energy to find the remaining potential energy:
\[ \text{PE}_f = \text{PE}_i - \text{KE} \] \[ \text{PE}_f = 5.80 , \text{kJ} - 4.61 , \text{kJ} \approx 1.19 , \text{kJ} \]
Now we will convert the final potential energy back into a height, using the formula for potential energy:
\[ \text{PE}_f = mgh_f \]
Solving for \( h_f \):
\[ h_f = \frac{\text{PE}_f}{mg} \]
Substituting in the known values:
\[ h_f = \frac{1190 , \text{J}}{118 , \text{kg} \times 9.81 , \text{m/s}^2} \]
Calculating \( h_f \):
\[ h_f = \frac{1190}{1155.78} \approx 1.03 , \text{m} \]
Thus, the height of the watermelon above the ground when its kinetic energy is 4.61 kJ is:
\[ h = h_i - h_f = 5.00 , \text{m} - 1.03 , \text{m} \approx 3.97 , \text{m} \]
So, the watermelon is approximately 3.97 meters above the ground when its kinetic energy is 4.61 kJ.
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