Final answer:
The spring constant for the spring launching the metal ball in this physics experiment is calculated by equating the spring's potential energy with the gravitational potential energy lost. After performing the calculations, the spring constant is found to be 29.43 N/m.
Step-by-step explanation:
To find the spring constant, we first need to determine the energy transferred to the metal ball. The gravitational potential energy gained by falling 1.5 meters can be calculated since this will equal the kinetic energy the ball had due to the launch from the spring when it left the spring's surface. Since the spring's potential energy is converted entirely into the kinetic energy of the ball (neglecting air resistance and other non-conservative forces), we can set the spring potential energy equal to the gravitational potential energy lost upon impact with the ground.
The gravitational potential energy (PE) lost is given by the formula PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity (9.81 m/s2), and h is the height fallen (1.5 m). For a 20 g ball, this would be:
PE = 0.020 kg * 9.81 m/s2 * 1.5 m = 0.2943 J (Joules)
The kinetic energy imparted to the ball by the spring (which gets converted to gravitational potential energy as it rises) is equal to the spring's potential energy, which is (1/2)kx2, where k is the spring constant and x is the compression distance of the spring.
So, we solve for k in the equation 0.2943 J = (1/2)k(0.20 m)2.
k = 2 * 0.2943 J / (0.20 m)2 = 29.43 N/m.
Therefore, the spring constant is 29.43 N/m.