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A 2.70-m-long pole is balanced vertically on its tip. It starts to fall and its lower end does not slip. Part A What will be the speed of the upper end of the pole just before it hits the ground? [Hint: Use conservation of energy.]

Respuesta :

Answer:

v = 7.27 m / s

Explanation:

We will use energy conservation at two points, the highest and the lowest

initial. Highest point

    Em₀ = U

    Em₀ = m g y

final. Lowest point

we set the reference system at this point so that the potential energy is zero

    [tex]Em_{f}[/tex] = K

   [tex]Em_{f}[/tex] = ½ m v²

how energy is conserved

      Em₀ = [tex]Em_{f}[/tex]

     mg y = ½ m v²

     v = √2 gy

     v = √ (2 9.8 2.7)

    v = 7.27 m / s

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