A 1.8 x 102-g particle is released from rest at point A on the inside of a smooth hemispherical bowl of radius R = 37.0 cm (figure below). (a) Calculate its gravitational potential energy at A relative to B. Energy of system at point A: PEA + KEA EA: PEA + KEA EA: PEA + 0 EA: PEA So, PEgravitational = mgh PEgravitational = (0.18)(9.81)(0.37) PEgravitational = 0.653 Got this one correct... (b) Calculate its kinetic energy at B. Energy of system at point B: PEB + KEB EB: PEB + KEB EB: mg(0) + KEB ***height is 0 at point B, so PE = 0*** EB: KEB The solution says that KEB = mgR Howwww????????!!!!!!!! Isn't it supposed to be (1/2)mv2??? Please help!