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A well-greased, essentially frictionless, metal bowl has the shape of a hemisphere of radius 0.150 m. You place a pat of butter of mass 5.00×10−3kg at the rim of the bowl and let it slide to the bottom of the bowl. What is the speed of the pat of butter when it reaches the bottom of the bowl?

Respuesta :

Answer:

[tex]v = 1.71 m/s[/tex]

Explanation:

As we know that there is no friction in the bowl so total mechanical energy is conserved here

So we can say that

initial potential energy of the butter = final kinetic energy at the bottom

so we can say

[tex]mgH = \frac{1}{2}mv^2[/tex]

here we can solve it for speed v

[tex]v^2 = 2gH[/tex]

now plug in all values in it

[tex]v^2 = (2 \times 9.8\times 0.150)[/tex]

[tex]v = 1.71 m/s[/tex]

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