Problem3 A 0.50 kg ball is attached to a 1.0 m long string and is hanging vertic
ID: 1777374 • Letter: P
Question
Problem3 A 0.50 kg ball is attached to a 1.0 m long string and is hanging vertically down on a string. We are holding the other end of the string. Suppose we now pull the ball out (with the string taut) and release it. At the bottom of the swing, the ball has a speed of 2.0 m/s. (a) What is the change in the ball's kinetic energy, from where we released it to the bottom of the swing? (b) How much total work was done on the ball (over the same interval)? (c) How much work does the string do on the ball? Hint-consider: what is the angle between the tension and the path being traveled by the ball at any instant? (d) Use the work kinetic energy theorem and your answers above to calculate how much work gravity does on the ball. (e ) The work done by gravity only depends on the change in the height and weight of the ball. Why is that true? (f) Use your answers to the questions above to answer the following question: how high did we raise the ball above its lowest position?Explanation / Answer
a) mass of the ball m = 0.5 kg
velocity v = 2.0m/s
KE = 0.5mv2 = 0.5*0.5*2.02 = 1 J
b) Total Work done = KE of the ball = 1J
c) Tension in the string is along its length. path of the ball is always perpendicular to the string, i.e tension. Hence work done by tension of the string is 0. ( W = F.ds)
d) There are only two forces Garvity and Tension
work done by Tension =0
work done by gravity = 1 J
e) change in height corrosponds to chnage in PE , hence work done by gravity is related to change in height. Gravitational force is vertical and height is vertical. The force is conservative . Hence work done depends only on change in height.
f) W = 1 J = mgh = 0.5*9.8*h
h = 2/9.8 = 0.20m
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