Respuesta :
Answer:v=10.84 m/s
Explanation:
Given
mass of Cylinder [tex]m=3 kg[/tex]
height of cylinder [tex]h=6 m[/tex]
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[tex]=mgh=3\times 9.8\times 6=176.4 J[/tex]
Kinetic Energy [tex]=\frac{1}{2}mv^2=\frac{1}{2}\times 3\times (v)^2[/tex]
[tex]176.4=\frac{1}{2}\times 3\times (v)^2[/tex]
[tex]v=\sqrt{117.6}[/tex]
[tex]v=10.84 m/s[/tex]
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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