A pendulum (assume no friction) is set up using a 0.125 kg ball at the end of a

Question

A pendulum (assume no friction) is set up using a 0.125 kg ball at the end of a string, such that the distance from the top of the string to the ball is 1.10 m. The ball is then pulled to the side, so that the string is horizontal.
a. The ball is released. What is the kinetic energy of the ball when it reaches the bottom of its swing?

b. What is the speed of the ball at the bottom of its swing
c. What is the kinetic energy of the ball when it reaches a vertical height of 35 cm on the way back up?

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Explanation / Answer

mass m = 0.125 kg distance from the top of the string to the ball is h = 1.10 m potential energy of the system after it is pulled ( i.e., in horizontal direction )PE = mgh = 1.3475 J (a).  the kinetic energy of the ball when it reaches the bottom of its swing               = potential energy of the system at horizontal position        KE = 1.3475 J Since from law of conservation of energy (b). we know KE = ( 1/ 2) mv^ 2 from this the speed of the ball at the bottom of its swing v = [2*KE / m]                                                                                        = 4.643 m / s (c).height h ' = 35 cm = 0.35 m the kinetic energy of the ball when it reaches a vertical height of 35 cm on the way back up is KE ' = KE - mgh '           = 1.3475 - 0.42875           = 0.91875 J

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