posted by Anonymous .
A smooth circular hoop with a radius of 0.400 m is placed flat on the floor. A 0.450-kg particle slides around the inside edge of the hoop. The particle is given an initial speed of 8.50 m/s. After one revolution, its speed has dropped to 4.00 m/s because of friction with the floor.
a) Find the energy transformed from mechanical to internal in the particle—hoop—floor system as a result of friction in one revolution.
b) What is the total number of revolutions the particle makes before stopping? Assume the friction force remains constant during the entire motion.
a) Relevant equation: Emec = DU + DK
If there are no nonconservative forces, this equation equals zero. Or in this case, some K is transformed into Eint (heat). So the equation becomes: Emec = Ff.d. (F in Newton, distance in meters and N.m = J).
-> DU + DK = Ff.d
Then, 0,5 m vi² = 0,5 m vf² + Ff.d
What do you need is: Ff.d = 0.5 m vi² - 05. m vf². (/!\ all the expression Ff.d is the answer, do not separate it).
b) You know what is the initial speed,
so you have the Ki = 0.5x0.4x64
Or you know that each turn you loose a certain quantity of energy (answer A).
-> Ki/ansA = ...