CHECKPOINT 1. What is the spring constant for a spring that starts at a length of 25 cm and extends to a length of 32 cm when a mass of 50g is added?

To find the spring constant (k) for a spring, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. The formula for Hooke's Law is:<br /><br />F = kx<br /><br />where F is the force applied to the spring, k is the spring constant, and x is the displacement of the spring from its equilibrium position.<br /><br />In this case, we are given the initial length of the spring (25 cm) and the final length of the spring (32 cm) when a mass of 50g is added. We can calculate the displacement (x) by subtracting the initial length from the final length:<br /><br />x = 32 cm - 25 cm = 7 cm<br /><br />Next, we need to calculate the force (F) exerted by the mass. The force exerted by a mass is given by the formula:<br /><br />F = mg<br /><br />where m is the mass and g is the acceleration due to gravity. In this case, the mass is 50g, which is equal to 0.05 kg, and the acceleration due to gravity is approximately 9.8 m/s^2. Plugging in these values, we get:<br /><br />F = (0.05 kg)(9.8 m/s^2) = 0.49 N<br /><br />Now that we have the force and the displacement, we can plug these values into Hooke's Law to solve for the spring constant:<br /><br />0.49 N = k(0.07 m)<br /><br />Dividing both sides of the equation by 0.07 m, we get:<br /><br />k = 0.49 N / 0.07 m = 7 N/m<br /><br />Therefore, the spring constant for the spring in this case is 7 N/m.