A m-kg bead is projected along the wire from A with speed u ms-1 so that it has enough energy to reach B but not C. Prove that 8.16<u<10.95 and that the speed at B >/= 2.58 ms-1. The vertical distance from (0, 0) to (3, 3) is $\displaystyle 3\sqrt{2}$ so the increase in

**potential energy** is $\displaystyle 3\sqrt{2}mg= 3\sqrt{2}(3)(9.8)= 153$ Joules. The vertical distance from (0, 0) to (6, 6) is $\displaystyle 6\sqrt{2}$ so the increase in potential energy is $\displaystyle 6\sqrt{2}mg= 6\sqrt{2}(3)(9.8)= 306$ Joules. Since the bead has enough

**kinetic energy**, initially, to go between B and C, what bounds must its speed be between? (I presume you know that kinetic energy is $\displaystyle (1/2)mv^2$.)