A spring of negligible mass is compressed by 0.250 m.Then a steel ball bearing of mass 0.05 kg is put on the spring,and the spring is released.The spring and ball bearing are aligned horizontally and friction is negligible.In addition,assume that the ball bearing will come off the spring exactly as the spring returns to its equilibrium position.The speed of the ball bearing right after the ball bearing loses contact with the spring is 12.9 m/s.What is the spring constant of the spring?

Question

Quizplus · Accepted Answer

The spring constant can be found using the conservation of energy principle. The potential energy stored in the spring when compressed is converted into kinetic energy of the ball bearing when the spring returns to its equilibrium position. The potential energy stored in the spring is given by $ \frac{1}{2}kx^2 $, where $ k $ is the spring constant and $ x $ is the compression distance. The kinetic energy of the ball bearing is given by $ \frac{1}{2}mv^2 $, where $ m $ is the mass of the ball bearing and $ v $ is its velocity. Setting these two energies equal to each other gives:$$ \frac{1}{2}kx^2 = \frac{1}{2}mv^2 $$Solving for $ k $ gives:$$ k = \frac{mv^2}{x^2} $$Plugging in the given values ($ m = 0.05 $ kg, $ v = 12.9 $ m/s, $ x = 0.250 $ m) gives:$$ k = \frac{0.05 \times (12.9)^2}{(0.250)^2} = 133 \, \text{N/m} $$

A Spring of Negligible Mass Is Compressed by 0