A uniform bar with seven evenly spaced holes is held by sliding the bar over a horizontal peg through one of the seven holes. The peg passes through hole C, and a cylinder of mass M hangs from a hook placed through hole A as shown above. The location of the center of mass of the bar is at the center of hole D. In this configuration, the bar-cylinder system remains motionless but is free to rotate around the peg in hole C. Frictional forces acting on the bar are negligible. In terms of the cylinder mass M, what is the mass of the bar? Briefly explain your answer.

The figure is missing but we can answer anyway.the correct answer is "releasing the string". in fact, when the person is holding the bow and the arrow, the kinetic energy of the arrow is still zero (because the arrow is not moving). the moment the arrow is released, it acquires a velocity v, so its kinetic energy will become different from zero: where m is the mass of the arrow and v its velocity.  all the other options, instead, do not increase the kinetic energy of the system: -  pulling farther back on the string : this will just increase the elastic potential energy of the system-  holding the arrow higher: this will increase the gravitational potential energy of the system (which is given by u=mgh, where h is the height above the ground)-  decreasing the mass of the arrow: this will decrease the gravitational potential energy of the system (which is given by u=mgh, where m is the mass)

It has more electrons than an neutral object