Question

conservation of mechanical energy quick check
a marble is attached to a compressed horizontal spring and is then released. which equation correctly shows the conservation of mechanical energy for this system? (1 point)
$mgh = kx^{2}$
$mgh = \frac{1}{2}kx^{2}$
$\frac{1}{2}mv^{2} = kx^{2}$
$\frac{1}{2}mv^{2} = \frac{1}{2}kx^{2}$

Explanation:

Step1: Analyze the initial and final energies

When the spring is compressed, the system has elastic potential energy \(E_{p}=\frac{1}{2}kx^{2}\) (where \(k\) is the spring - constant and \(x\) is the compression of the spring). Since the spring is horizontal, there is no change in gravitational potential energy (\(mgh = 0\) as \(h\) is constant). When the marble is released, the elastic potential energy is converted into kinetic energy \(E_{k}=\frac{1}{2}mv^{2}\) (where \(m\) is the mass of the marble and \(v\) is its velocity).

Step2: Apply the conservation of mechanical energy

By the law of conservation of mechanical energy \(E_{initial}=E_{final}\). The initial energy is the elastic potential energy of the spring and the final energy is the kinetic energy of the marble. So, \(\frac{1}{2}mv^{2}=\frac{1}{2}kx^{2}\)

Answer:

\(\frac{1}{2}mv^{2}=\frac{1}{2}kx^{2}\) (the fourth option)