A 4-kg block sits on top of a shelf. Connected by a light and frictionless string to the 4-kg block, a 2-kg block is suspended over a pulley on the edge of the shelf. The coefficient of kinetic friction between the 4-kg block and the shelf is 0.35.
a)Find the energy dissipated by friction when the 2-kg block falls a distance y
b)Find the total mechanical energy E of the two-block system after the 2-kg block falls a distance y
c)Use your result for (b) to find the speed of either block after the 2-kg block falls 2 m
This is in relation to Kinetic and Potential Energy. I've racked up my brain all night to solve this...Help!
2007-01-28 04:43:09 · 1 answers · asked by bl00ish 2 in Science & Mathematics ➔ Physics
not Kinetic and Potential Energy for part a. Just the old E = Fd
a) Using E = Fd, you are given the distance y, well okay its still a variable but you are given it as y. The force F is just the weight of the 4kg block which is 4 * 9.8 = 39.2nt times the coefficient of friction or 39.2 * .35 = 13.7newtons. So E = 13.7y
b) Using Potential Energy, E = hmg using h = y (really just the change in potential energy), m = 2kg and g = 9.8 we get E = 19.6y. The only issue might be the sign of the Energy. We know that the potential energy is getting less as objects fall so maybe a -19.6y
c) Now we have the potential energy minus the frictional energy or (19.6 - 13.7)y = 5.9y . This would be the Kinetic Energy that you are looking for so you now have 1/2mv^2 = 5.9y set y to 2 meters and m = (4kg + 2kg) = 6kg then you can solve for v.
2007-01-28 05:02:51 · answer #1 · answered by rscanner 6 · 0⤊ 0⤋