A mass of 2.0 kg traveling at 3.0 m/s along a smooth, horizontal plane hits a relaxed spring. The mass is slowed to zero velocity when the spring has been compressed by 0.15 m. What is the spring constant of the spring?

Question

Quizplus · Accepted Answer

The correct answer is A because the spring constant is given by the equation k = F/x, where F is the force applied to the spring and x is the displacement of the spring. In this case, the force applied to the spring is equal to the mass of the object times the acceleration of the object, which is given by F = ma. The acceleration of the object is equal to the change in velocity divided by the time it takes to change velocity, which is given by a = (v_f - v_i)/t. In this case, the initial velocity of the object is 3.0 m/s, the final velocity of the object is 0 m/s, and the time it takes to change velocity is equal to the time it takes for the spring to be compressed by 0.15 m, which is given by t = x/v. In this case, the time it takes for the spring to be compressed by 0.15 m is equal to 0.15 m / 3.0 m/s = 0.05 s. Therefore, the acceleration of the object is equal to (0 m/s - 3.0 m/s) / 0.05 s = -60 m/s^2. Therefore, the force applied to the spring is equal to (2.0 kg)(-60 m/s^2) = -120 N. Therefore, the spring constant is equal to (-120 N) / (0.15 m) = 800 N/m.

A Mass of 2