Question

A 3.0-kg cylinder falls vertically from rest in a very tall, evacuated tube near the surface of the earth. What is its speed after the cylinder has fallen 6.0 m?

Answer 1

Answer:v=10.84 m/s

Explanation:

Given

mass of Cylinder $m=3 kg$

height of cylinder $h=6 m$

It is given that tube is evacuated so we can neglect air resistance so friction provided by the air is zero.

Since Energy cannot be destroyed but can be transformed from one form to another therefore Potential Energy of Cylinder is converted to Kinetic Energy of Cylinder

Potential Energy$=mgh=3\times 9.8\times 6=176.4 J$

Kinetic Energy $=\frac{1}{2}mv^2=\frac{1}{2}\times 3\times (v)^2$

$176.4=\frac{1}{2}\times 3\times (v)^2$

$v=\sqrt{117.6}$

$v=10.84 m/s$

Answer 2

Speed of cylinder after the cylinder has fallen 6 m is 11 meter/second.

Energy conservation based problem

What information do we have?

Mass of cylinder = 3 kg

Height = 6 m

Using energy conservation theroy

mgh = (1/2)mv²

gh = (1/2)v²

(9.8)(6) = (1/2)v²

Velocity = 11 m/s (Approx.)

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