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A simple pendulum consists of a 4.9-kg mass attached to a string. The pendulum is pulled to the right and held at rest so that it is 2.08 m above the lowest point in its swing. If you let it go from this point, how fast (in m/s) is it traveling at the lowest point of its swing (on the initial pass)

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pmdislamabad

Answer:

6.384 m/[tex]sec^{2}[/tex]

Explanation:

The velocity is given by;

We know;

PE= mgh

KE= 1/2 m[tex]V^{2}[/tex]

but KE=PE

==> 1/2 m[tex]V^{2}[/tex] =mgh

==> 1/2 [tex]V^{2}[/tex] =gh

==> [tex]v=\sqrt{2gh}[/tex]

puting g=9.8 m/sec2

h=2.08 m

==> [tex]v= 6.384 m/sec^{2}[/tex]=

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