An archer draws a bow, getting ready to shoot a 55.0 gram arrow. The tension in the bowstring is 424 N, and it is bent at the angles 30 degrees

Question

a)What is the magnitude of the "drawing" force holding the string at rest?

b)If the drawing force is removed (by the archer releasing the string to shoot the arrow), what will be the acceleration of the arrow? All forces can be neglected on the arrow except the forces being exerted by the bowstring.

Answer 1

F = 2 * 424 * sin 30 = 424

F = m A

A = 424/.055 = 7709 m/s^2