ch falling object has the least kinetic energy when it collides on the ground?(1 point)

Responses

120 Kg of sand falling 12 m/s
120 Kg of sand falling 12 m/s

12 Kg of sand falling at 8 m/s
12 Kg of sand falling at 8 m/s

12 Kg of water falling at 10 m/s
12 Kg of water falling at 10 m/s

120 Kg of water falling at 12 m/s

1 answer

Kinetic energy (KE) can be calculated using the formula:

\[ \text{KE} = \frac{1}{2} mv^2 \]

where \( m \) is the mass and \( v \) is the velocity of the object.

Let's calculate the kinetic energy for each case:

  1. 120 Kg of sand falling at 12 m/s \[ \text{KE} = \frac{1}{2} \times 120 , \text{kg} \times (12 , \text{m/s})^2 = \frac{1}{2} \times 120 \times 144 = 8640 , \text{J} \]

  2. 12 Kg of sand falling at 8 m/s \[ \text{KE} = \frac{1}{2} \times 12 , \text{kg} \times (8 , \text{m/s})^2 = \frac{1}{2} \times 12 \times 64 = 384 , \text{J} \]

  3. 12 Kg of water falling at 10 m/s \[ \text{KE} = \frac{1}{2} \times 12 , \text{kg} \times (10 , \text{m/s})^2 = \frac{1}{2} \times 12 \times 100 = 600 , \text{J} \]

  4. 120 Kg of water falling at 12 m/s \[ \text{KE} = \frac{1}{2} \times 120 , \text{kg} \times (12 , \text{m/s})^2 = \frac{1}{2} \times 120 \times 144 = 8640 , \text{J} \]

Comparing the kinetic energies:

  • 120 Kg of sand at 12 m/s = 8640 J
  • 12 Kg of sand at 8 m/s = 384 J
  • 12 Kg of water at 10 m/s = 600 J
  • 120 Kg of water at 12 m/s = 8640 J

The object with the least kinetic energy is 12 Kg of sand falling at 8 m/s, with a kinetic energy of 384 J.