Avertical spring has a spring constant of 100 Newtons per meter. When an object is attached to the bottom of the spring, the spring changes from its unstretched length of 0.50 meter to a length of 065 meter. The magnitude of the weight of the attached object is

## Step 1:<br />This problem is related to Hooke's law, which states that the force needed to extend or compress a spring by some distance \(x\) is proportional to that distance. The mathematical representation of Hooke's law is \(F=kx\), where \(F\) is the force applied on the spring or by the spring when it is compressed or extended, \(k\) is the spring constant, and \(x\) is the distance the spring is stretched or compressed.<br /><br />## Step 2:<br />From the problem, we know that the spring constant \(k = 100\) Newtons per meter, the uncompressed or initial length of the spring is \(0.50\) m, and when the object is attached to the spring, the length of the spring extends to \(0.65\) m.<br /><br />## Step 3:<br />We calculate the amount the spring is stretched:<br /><br />### \(x = \text{Final Length} - \text{Initial Length} = 0.65 \, m - 0.50 \, m = 0.15 \, m\)<br /><br />## Step 4:<br />We substitute the given values into Hooke's law:<br /><br />### \(F = kx = 100 \, N/m \times 0.15 \, m = 15 \, Newtons\)