Question

potential and kinetic energy. the rope that is being used to hang the rock is going to be cut. when the rock is falling, at what height are the potential and kinetic energy of the rock almost equal? 3 m 2 m 1 m 0 m meters

Explanation:

Step1: Recall energy - conservation formula

The total mechanical energy \(E = mgh+\frac{1}{2}mv^{2}\), where \(m\) is the mass of the rock, \(h\) is the height, \(g\) is the acceleration due to gravity (\(g\approx9.8m/s^{2}\)), and \(v\) is the velocity. Initially, when the rock is at rest and hanging, its total energy is potential energy \(E = mgh_0\) (where \(h_0\) is the initial height). Let the initial height \(h_0 = 3m\).

Step2: Set potential and kinetic energy equal

When \(U = K\), we know that \(U=mgh\) and \(K=\frac{1}{2}mv^{2}\), and from energy - conservation \(mgh_0=mgh+\frac{1}{2}mv^{2}\). Since \(mgh=\frac{1}{2}mv^{2}\), then \(mgh_0 = 2mgh\).

Step3: Solve for the height \(h\)

Dividing both sides of the equation \(mgh_0 = 2mgh\) by \(mg\) (since \(m
eq0\) and \(g
eq0\)), we get \(h=\frac{h_0}{2}\). Given \(h_0 = 3m\), then \(h = 1.5m\). But among the given options, the closest value is \(1m\).

Answer:

1