potential and kinetic energy\nthe rope that is being used to hang the rock is going to be cut. when the rock…

potential and kinetic energy\nthe 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?\n3 m\n2 m\n1 m\n0 m

potential and kinetic energy\nthe 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?\n3 m\n2 m\n1 m\n0 m

Answer

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 above the ground, (g) is the acceleration due to gravity ((g\approx9.8m/s^{2})), and (v) is the velocity of the rock. Initially, when the rock is at rest just before the rope is cut, (E = mgh_0) (where (h_0) is the initial height, here (h_0 = 3m)). When the potential energy (U = mgh) and kinetic energy (K=\frac{1}{2}mv^{2}) are equal, (U = K). Also, from energy - conservation (E=U + K). Since (U = K), then (E = 2U).

Step2: Substitute and solve for (h)

We know (E=mgh_0) and (E = 2mgh). So, (mgh_0=2mgh). Canceling out the mass (m) and the acceleration due to gravity (g) on both sides of the equation, we get (h=\frac{h_0}{2}). Given (h_0 = 3m), then (h = 1.5m). Since the closest value among the given options is (2m).

Answer:

2 m