To determine which falling object has the least kinetic energy when it collides with the ground, we can use the formula for kinetic energy (KE):
\[ KE = \frac{1}{2} mv^2 \]
where \(m\) is the mass of the object and \(v\) is its velocity.
Let's calculate the kinetic energy for each of the given objects:
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120 Kg of sand falling at 12 m/s: \[ KE = \frac{1}{2} \times 120 , kg \times (12 , m/s)^2 = \frac{1}{2} \times 120 \times 144 = 8640 , J \]
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12 Kg of sand falling at 8 m/s: \[ KE = \frac{1}{2} \times 12 , kg \times (8 , m/s)^2 = \frac{1}{2} \times 12 \times 64 = 384 , J \]
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12 Kg of water falling at 10 m/s: \[ KE = \frac{1}{2} \times 12 , kg \times (10 , m/s)^2 = \frac{1}{2} \times 12 \times 100 = 600 , J \]
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120 Kg of water falling at 12 m/s: \[ KE = \frac{1}{2} \times 120 , kg \times (12 , m/s)^2 = \frac{1}{2} \times 120 \times 144 = 8640 , J \]
Now, we summarize the kinetic energies calculated:
- 120 Kg of sand falling at 12 m/s: 8640 J
- 12 Kg of sand falling at 8 m/s: 384 J
- 12 Kg of water falling at 10 m/s: 600 J
- 120 Kg of water falling at 12 m/s: 8640 J
From these calculations, we can see that the object with the least kinetic energy is 12 Kg of sand falling at 8 m/s, with a kinetic energy of 384 J.