An amiount of work equal to 1.5 J is required to compress the spring in a spring-gun. What is the "launch speed" of a \( 15-\mathrm{g} \) marble? 1) \( 14 \mathrm{~m} / \mathrm{s} \) 2) \( 15 \mathrm{~m} / \mathrm{s} \) 3) \( 18 \mathrm{~m} / \mathrm{s} \) 4) \( 21 \mathrm{~m} / \mathrm{s} \) 5) \( 23 \mathrm{~m} / \mathrm{s} \)

To find the launch speed of the marble, we can use the principle of conservation of energy. The work done on the spring (1.5 J) is converted into kinetic energy when the marble is launched. The kinetic energy (KE) can be expressed as: \[ KE = \frac{1}{2}mv^2 \] where \( m \) is the mass of the marble and \( v \) is the speed. First, we need to convert the mass from grams to kilograms: \[ m = 15 \, \text{g} = 0.015 \, \text{kg} \] Now, we can set the work done equal to the kinetic energy: \[ 1.5 \, \text{J} = \frac{1}{2}(0.015 \, \text{kg})v^2 \] By rearranging the equation to solve for \( v \): \[ v^2 = \frac{2 \times 1.5 \, \text{J}}{0.015 \, \text{kg}} = 200 \] Taking the square root gives us: \[ v = \sqrt{200} \approx 14.14 \, \text{m/s} \] Therefore, rounded to the nearest whole number, the launch speed of the marble is approximately **14 m/s**, corresponding to option 1. 1) Throughout history, springs have been ingeniously applied in various mechanisms from ancient catapults to modern toys. The physics of springs, explained by Hooke's law, describes how the force exerted is directly proportional to their compression or extension—a principle harnessed for both fun and complex engineering solutions! 2) In real-world applications, spring guns illustrate the transformation of potential energy into kinetic energy, a concept seen in many mechanical systems like watches and vehicle suspensions. Understanding this energy conversion is essential not only in scientific studies but also in designing efficient energy management systems across various industries.