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a 3.0 - kg cylinder falls vertically from rest near the surface of the earth. air resistance is negligible. use the conservation of total energy to find the speed of the cylinder after it has fallen 6.0 m?
120 m/s
13 m/s
8.0 m/s
26 m/s
11 m/s

Explanation:

Step1: Apply conservation of energy

The initial energy \(E_{i}\) is potential energy \(E_{i}=mgh\) (since it starts from rest, kinetic energy \(K_{i} = 0\)). The final energy \(E_{f}\) is kinetic energy \(E_{f}=\frac{1}{2}mv^{2}\) (potential energy at the lower - level \(h = 0\)). By the conservation of energy \(E_{i}=E_{f}\), so \(mgh=\frac{1}{2}mv^{2}\).

Step2: Solve for \(v\)

Cancel out the mass \(m\) from both sides of the equation \(mgh=\frac{1}{2}mv^{2}\). We get \(gh=\frac{1}{2}v^{2}\). Then \(v=\sqrt{2gh}\). Given \(g = 9.8\ m/s^{2}\) and \(h = 6.0\ m\), substitute the values: \(v=\sqrt{2\times9.8\times6.0}=\sqrt{117.6}\approx 11\ m/s\).

Answer:

11 m/s