A block of mass 0.249 kg is placed on top of a light, vertical spring of force constant 5 175 N/m and pushed downward so that the spring is compressed by 0.105 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise? (Round your answer to two decimal places.) X Where is energy stored before the block is released? Where is energy stored when the block reaches its maximum height? m Need Help? Read ItTutorial Exercise A bead with a hole through it slides on a wire track. The wire is threaded through the hole in the bead, and the bead slides without friction around a loop-the-loop (see figure below). The bead is released from rest at a height h = 3.45R. A (a) What is its speed at point (47 (b) How large is the normal force on the bead at point @ if its mass is 4.55 g? Part 1 of 4 - Conceptualize Since the bead is released above the top of the loop, it starts with enough potential energy to reach point @ and still have excess kinetic energy. The energy of the bead at point @ is proportional to h and g. If it is moving relatively slowly, the track will exert an upward force on the bead, but if it is whipping around quickly, the normal force will push it toward the center of the loop. Part 2 of 4 - Categorize The speed at the top of the loop can be found from the conservation of energy for the bead-track-Earth system and the normal force can be found from Newton's second law. Part 3 of 4 - Analyze (a) We define the bottom of the loop as the zero level for the gravitational potential energy. For the total energy of the system at point @, we have Uj + K ; = Uf+ KF. Now we know that U. = mgh and K; = 0 because the bead starts from rest at height h. At point @, the bead is at height R with speed v, so we have mgh + 0 = mg R ) + Amv2. Substituting the given expression for h in terms of R, gives mg R) = mg R ) + Amv2. Now we can solve for v in terms of R and g. 2 = 9 R 1 9R