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A 2.0m long pendulum is released from rest when the support string is at an angle of 25 degrees with the vertical. What is the speed of the bob at the bottom of the swing?

Respuesta :

Khoso123

Answer:

[tex]v=1.92m/s[/tex]

Explanation:

Given data

Length h=2.0m

Angle α=25°

To find

Speed of bob

Solution

From conservation of energy we know that:

[tex]P.E=K.E\\mgh=(1/2)mv^{2}\\ gh=(1/2)v^{2}\\v^{2}=\frac{gh}{0.5}\\ v=\sqrt{\frac{gh}{0.5}}\\ v=\sqrt{\frac{(9.8m/s^{2} )(2.0-2.0Cos(25^{o} ))}{0.5}}\\v=1.92m/s[/tex]

Cricetus

The speed of the bob will be "1.92 m/s".

Given values:

  • Length, h = 2.0 m
  • Angle, α = 25°

As we know,

The conservation of energy:

→ [tex]Potential \ energy = Kinetic \ energy[/tex]

or,

→ [tex]mgh = \frac{1}{2} mv^2[/tex]

or,

→    [tex]v^2 = \frac{gh}{0.5}[/tex]

          [tex]= \sqrt{\frac{gh}{0.5} }[/tex]

By substituting the values, we get

          [tex]= \sqrt{\frac{(9.8)(2.0-2.0 Cos (25^{\circ}))}{0.5} }[/tex]

          [tex]= 1.92 \ m/s[/tex]

Thus the above answer is right.

Learn more about speed here:

https://brainly.com/question/7247767

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