Professor Flitwick pushed his 1.5 kg Charms textbook against a 5 N/m spring causing it to compress leftwards 1.5 m over a rough spot on the table that has a coefficient of friction of 0.25. He then released the textbook, so it launches 2.7 m across the 0.9 m high table and flies off at 70 degrees to the horizontal. If there is a group of students 0.95 m away from the base of the table, will the textbook land in front of them or will this be slightly embarrassing for Flitwick?

If the spring and table are horizontal, the energy lost to friction during spring expansion is W = Fd = μmgd = 0.25(1.5)(9.8)(1.5) = 5.5125 J so only a tiny amount of kinetic energy (5.625 - 5.5125 = 0.1125 J) exists after spring contact is lost.

If the remaining portion of the 2.7 m tabletop has friction, the book stops fairly quickly after departing the spring. If the tabletop is otherwise frictionless, the book slides horizontally off the table with a velocity of

0.1125 = ½(1.5)v²....v = 0.378 m/s

from vertical rest, the book takes t = √(2(0.9)/9.8) = 0.429 s to hit the floor. At floor contact, the book is d = vt = 0.378(0.429) = 0.162 m from the table. It does not hit the students. This analysis does not take into account the 70° angle in any way.

If we assume a frictionless ramp that converts the horizontal motion to motion at + 70° with no loss of launch velocity (haha). Then the time to strike the floor comes from the formula

0 = 0.9 + (0.378sin70)t + ½(-9.8)t²

0 = 0.9 + 0.355t - 4.9t²

quadratic formula positive answer

t = (-0.355 - √(0.355² - 4(-4.9)(0.9))) / -9.8 = 0.466 s

and the book lands

d = vt = (0.378cos70)(0.466) = 0.06 m from the base of the table.