A small block with a mass of 0.0600 kg isattached to a cord passing through a ho
ID: 1783538 • Letter: A
Question
A small block with a mass of 0.0600 kg isattached to a cord passing through a hole in a frictionless, horizontal surface. The block is originallyrevolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s. The cord is then pulledfrom below, shortening the radius of the circle in which the block revolves to 0.10 m. At this newdistance, the speed of the block is 2.80 m/s.
d) Now let’s calculate the work usingW= F·dl. To do this you will need a result that comesfrom angular momentum conservation, namely that the velocity will change as1/R. For theproblem as stated,v= 0.28/R, allowing you to determineT(R).
Explanation / Answer
K = W so
K2 = ½ mv22 = 0.5 * 0.06kg*2.8^2 = 0.2352 J
K1 = ½ mv12 = 0.5 * 0.06kg*0.7^2 = 0.0147 J
k = K2 - K1 = 0.2352 - 0.0147 = 0.2205 J