To calculate the kinetic energy (KE) of the suitcase just before it hits the floor, you can use the kinetic energy formula:
\[ KE = \frac{1}{2} m v^2 \]
where:
- \( m \) is the mass of the suitcase (in kilograms),
- \( v \) is the velocity (in meters per second).
Given:
- \( m = 14 , \text{kg} \)
- \( v = 2.7 , \text{m/s} \)
Now, plug in the values:
\[ KE = \frac{1}{2} \times 14 , \text{kg} \times (2.7 , \text{m/s})^2 \]
Calculating \( (2.7 , \text{m/s})^2 \):
\[ (2.7)^2 = 7.29 , \text{m}^2/\text{s}^2 \]
Now substitute this back into the equation:
\[ KE = \frac{1}{2} \times 14 , \text{kg} \times 7.29 , \text{m}^2/\text{s}^2 \]
Calculating:
\[ KE = 7 , \text{kg} \times 7.29 , \text{m}^2/\text{s}^2 \] \[ KE = 51.03 , \text{kg} \cdot \text{m}^2/\text{s}^2 \]
Since \( 1 , \text{kg} \cdot \text{m}^2/\text{s}^2 = 1 , \text{J} \), we have:
\[ KE \approx 51.03 , \text{J} \]
Thus, the kinetic energy of the suitcase just before it hits the floor is approximately 51.03 joules.