Question

Which falling object will result in the least kinetic energy when it collides with the ground?(1 point) Responses 10 grams of dirt falling at 2 m/s 10 grams of dirt falling at 2 m/s 100 grams of dirt falling at 15 m/s 100 grams of dirt falling at 15 m/s 10 grams of water falling at 1 m/s 10 grams of water falling at 1 m/s 100 grams of water falling at 10 m/s

Answer 1

To determine which falling object results in the least kinetic energy when it collides with the ground, we can use the formula for kinetic energy (KE):

\[
KE = \frac{1}{2} m v^2
\]

where \( m \) is the mass in kilograms and \( v \) is the velocity in meters per second.

Now, let's calculate the kinetic energy for each of the options:

1. **10 grams of dirt falling at 2 m/s**:
- Convert mass to kilograms: \( 10 \text{ g} = 0.01 \text{ kg} \)
- Calculate KE:
\[
KE = \frac{1}{2} (0.01 \text{ kg})(2 \text{ m/s})^2 = \frac{1}{2} (0.01)(4) = 0.02 \text{ J}
\]

2. **100 grams of dirt falling at 15 m/s**:
- Convert mass to kilograms: \( 100 \text{ g} = 0.1 \text{ kg} \)
- Calculate KE:
\[
KE = \frac{1}{2} (0.1 \text{ kg})(15 \text{ m/s})^2 = \frac{1}{2} (0.1)(225) = 11.25 \text{ J}
\]

3. **10 grams of water falling at 1 m/s**:
- Convert mass to kilograms: \( 10 \text{ g} = 0.01 \text{ kg} \)
- Calculate KE:
\[
KE = \frac{1}{2} (0.01 \text{ kg})(1 \text{ m/s})^2 = \frac{1}{2} (0.01)(1) = 0.005 \text{ J}
\]

4. **100 grams of water falling at 10 m/s**:
- Convert mass to kilograms: \( 100 \text{ g} = 0.1 \text{ kg} \)
- Calculate KE:
\[
KE = \frac{1}{2} (0.1 \text{ kg})(10 \text{ m/s})^2 = \frac{1}{2} (0.1)(100) = 5 \text{ J}
\]

Now, let's summarize the kinetic energies calculated:

- **10 grams of dirt at 2 m/s**: 0.02 J
- **100 grams of dirt at 15 m/s**: 11.25 J
- **10 grams of water at 1 m/s**: 0.005 J
- **100 grams of water at 10 m/s**: 5 J

The object with the least kinetic energy when it collides with the ground is **10 grams of water falling at 1 m/s**, with a kinetic energy of **0.005 J**.