Kinetic energy (KE) can be calculated using the formula:
\[ KE = \frac{1}{2} m v^2 \]
where:
- \( m \) is the mass of the object (in kilograms),
- \( v \) is the velocity of the object just before it hits the ground (in meters per second).
To determine which object generates the most kinetic energy before hitting the ground, we need to figure out the velocity of each object just before it hits the ground. Assuming they all fall from the same height, we can calculate their final velocity by using the equations of motion under gravity or simply knowing that the potential energy will be converted to kinetic energy.
To find the speed (v) of a falling object from a height \( h \), you could use the formula derived from energy conservation:
\[ v = \sqrt{2gh} \]
where \( g \) is approximately \( 9.81 , m/s^2 \) (acceleration due to gravity). However, let’s simplify the comparison by directly using the masses and their velocities provided.
Let's calculate the kinetic energy for each object:
-
For 110 kg at 10 m/s: \[ KE = \frac{1}{2} \times 110 \times 10^2 = \frac{1}{2} \times 110 \times 100 = 5500 \text{ J} \]
-
For 75 kg at 2 m/s: \[ KE = \frac{1}{2} \times 75 \times 2^2 = \frac{1}{2} \times 75 \times 4 = 150 \text{ J} \]
-
For 50 kg at 1 m/s: \[ KE = \frac{1}{2} \times 50 \times 1^2 = \frac{1}{2} \times 50 \times 1 = 25 \text{ J} \]
-
For 10 kg at 10 m/s: \[ KE = \frac{1}{2} \times 10 \times 10^2 = \frac{1}{2} \times 10 \times 100 = 500 \text{ J} \]
Now summarizing the kinetic energy we calculated:
- 110 kg at 10 m/s: 5500 J
- 75 kg at 2 m/s: 150 J
- 50 kg at 1 m/s: 25 J
- 10 kg at 10 m/s: 500 J
Conclusion: The object with a mass of 110 kg dropped from the same height and having a speed of 10 m/s will generate the most kinetic energy before hitting the ground, which is 5500 J.