C Not Secure l wwwwebassigr.net/web/StudentiAssignmer potential energy to reach
ID: 776641 • Letter: C
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C Not Secure l wwwwebassigr.net/web/StudentiAssignmer potential energy to reach point Since the beed is released above the top of the loop, it starts and still have excess kinetic energy. The energy of the bead at point pls proportional to h and g. If 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-Categortze 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 Now we know that U.-mgh and K-O because the bead starts from rest at height h. At point is at height 3.05 R with speed v, so we have , the bead mph … mg(3.05 R)+3m2. - Substituting the given expression for h in terms of R, gives Now we can solve for v in terms of R and g gR, Need Help? Raadit Submit Answer Save Progress Practice Another Version -/3 points SerPSET9 8.P.006 A block of mass m-4.30 kg is released from rest from point O and slides on the frictionless track shown in the figure belo MacBook AirExplanation / Answer
a)
mgh = mgR + (0.5) m v2
h = 3.05 R
hence
(3.05)mgR = mgR + (0.5) m v2
(2.05) mgR = (0.5) m v2
v2 = 4.1 gR
v = sqrt(4.1 gR)
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