A horizontal spring having constant k = 50.0 N/m is attached to a block having mass 0.50 kg.The mass is oscillating freely on a frictionless table with amplitude 15.0 cm.As the mass travels from the position of maximum stretch to half the position of maximum stretch,how much work is done by the spring on the mass?

Question

Quizplus · Accepted Answer

The work done by the spring on the mass can be calculated using the change in potential energy of the spring system, which is given by $ \Delta U = \frac{1}{2}k(x_2^2 - x_1^2) $, where $x_1$ and $x_2$ are the initial and final displacements from the equilibrium position, respectively. Here, $x_1 = 15.0 \, \text{cm} = 0.15 \, \text{m}$ (maximum stretch) and $x_2 = 0.5 \times 0.15 \, \text{m} = 0.075 \, \text{m}$ (half the maximum stretch). Plugging in the values, we get $ \Delta U = \frac{1}{2} \times 50.0 \, \text{N/m} \times (0.075^2 - 0.15^2) \, \text{m}^2 = +0.42 \, \text{J} $. Positive work indicates the spring is doing work on the mass as it moves towards the equilibrium position.

A Horizontal Spring Having Constant K = 50