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A 2.00 Kg pendulum bob on a string 1.20 m long is released with a velocity of 3.00 m/s when the support string makes an angle of 30.0 degrees with the vertical. What is the velocity of the swing?A. 3.49 m/sB. 4.00 m/sC. 4.32 m/sD. 3.75 m/s

Question

A 2.00 Kg pendulum bob on a string 1.20 m long is released with a velocity of 3.00 m/s when the support string makes an angle of 30.0 degrees with the vertical. What is the velocity of the swing?

A. 3.49 m/s

B. 4.00 m/s

C. 4.32 m/s

D. 3.75 m/s

check_circleAnswer
Step 1

According to the conservation of energy, the total energy at a height is equal to the total energy at the bottom of the swing.

K,+U K+U
mu2 mgh=-mv2 +0
2
2
+mgl (1-cose)
2
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Image Transcriptionclose

K,+U K+U mu2 mgh=-mv2 +0 2 2 +mgl (1-cose) 2

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Step 2

Substitute the values in the above relati...

(2.00)(3.00)(2.00) (9.80)(1.20) (1 -cos30) -(2.00)
9.0+3.15=
12.15
v 3.49 m/s
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(2.00)(3.00)(2.00) (9.80)(1.20) (1 -cos30) -(2.00) 9.0+3.15= 12.15 v 3.49 m/s

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