To determine which falling object will result in the least kinetic energy when it collides with the ground, we can use the kinetic energy formula:
\[ KE = \frac{1}{2} m v^2 \]
where \( KE \) is kinetic energy, \( m \) is mass in kilograms, and \( v \) is velocity in meters per second.
Now, let's calculate the kinetic energy for each option:
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10 grams of dirt falling at 2 m/s:
- Mass \( m = 10 , \text{g} = 0.01 , \text{kg} \)
- Velocity \( v = 2 , \text{m/s} \)
- \( KE = \frac{1}{2} (0.01) (2^2) = \frac{1}{2} (0.01) (4) = 0.02 , \text{J} \)
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10 grams of water falling at 1 m/s:
- Mass \( m = 10 , \text{g} = 0.01 , \text{kg} \)
- Velocity \( v = 1 , \text{m/s} \)
- \( KE = \frac{1}{2} (0.01) (1^2) = \frac{1}{2} (0.01) (1) = 0.005 , \text{J} \)
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100 grams of dirt falling at 15 m/s:
- Mass \( m = 100 , \text{g} = 0.1 , \text{kg} \)
- Velocity \( v = 15 , \text{m/s} \)
- \( KE = \frac{1}{2} (0.1) (15^2) = \frac{1}{2} (0.1) (225) = 11.25 , \text{J} \)
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100 grams of water falling at 10 m/s:
- Mass \( m = 100 , \text{g} = 0.1 , \text{kg} \)
- Velocity \( v = 10 , \text{m/s} \)
- \( KE = \frac{1}{2} (0.1) (10^2) = \frac{1}{2} (0.1) (100) = 5 , \text{J} \)
Now, compare the kinetic energies:
- 10 grams of dirt at 2 m/s: \( 0.02 , \text{J} \)
- 10 grams of water at 1 m/s: \( 0.005 , \text{J} \)
- 100 grams of dirt at 15 m/s: \( 11.25 , \text{J} \)
- 100 grams of water at 10 m/s: \( 5 , \text{J} \)
The object that results in the least kinetic energy when it collides with the ground is 10 grams of water falling at 1 m/s with \( 0.005 , \text{J} \).
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