A \( 7.00-\mathrm{kg} \) bowling ball falls from a \( 2.00-\mathrm{m} \) shelf. Just before hitting the floor, what will be its kinetic energy? \( \left(g=9.80 \mathrm{~m} / \mathrm{s}^{2}\right. \) and assume air resistance is negligible) 1) 14.0 J 2) 19.6 J 3) 29.4 J 4) 137 J 5) 156 J

To find the kinetic energy just before the bowling ball hits the floor, we can use the concept of gravitational potential energy (PE), which converts to kinetic energy (KE) as it falls. The PE at the height of 2.00 m is given by the formula: \[ PE = mgh \] Where: - \( m = 7.00 \, \text{kg} \) - \( g = 9.80 \, \text{m/s}^2 \) - \( h = 2.00 \, \text{m} \) Calculating this: \[ PE = 7.00 \, \text{kg} \times 9.80 \, \text{m/s}^2 \times 2.00 \, \text{m} = 137.2 \, \text{J} \] Since we assume air resistance is negligible, this potential energy converts entirely to kinetic energy just before impact. Thus, the kinetic energy just before hitting the floor is approximately 137 J. So, the correct answer is **4) 137 J**. Now, did you know that bowling has a rich history dating back thousands of years? It originated in ancient societies like Egypt where excavations have revealed balls and pins—what a throwback to the original "strike"! And, if you ever feel tempted to bowl a perfect game, remember to pay attention to your grip and follow-through. Many bowlers mistakenly focus too much on speed, but it's all about making that smooth connection with the ball. Happy bowling!