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Help please www.webassign.net/web/Student/Assignment-Responses/submit?dep-16772170 120% c asearch Need Help? iWatch Read It 3.i 1/3 points ! Previous Answers serPSE9 8 P 005 MI A bead slides without friction around a loop-the-loop (see figure below). The bead is released from rest at a height h 3.30R. (Use the following as necessary: the acceleration due to gravity g, and R.) (o) What is its speed at point - 2.6 gr (b) How large is the normal force on the bead at point if its mass is 5.20 magnitude 01834 N direction downward grams Need Help?Master

Explanation / Answer

a] At A, the change in Potential Energy will be:

mgh - mg(2R)

by conservation of energy, this will be equal to kinetic energy at A

so, gh - 2gR = (1/2)v2

=> v = [2gh - 4gR]1/2

=> v = [6.6gR - 4gR]1/2

=> v = [2.6gR]1/2

this is the speed at A.

b] The net force at A will be:

mg + P = mv2/R

Where P is the normal force

=> P = m[2.6g] - mg = [1.6mg]

so, P = 1.6 x 5.2 x 10-3 x 9.8 = 0.0815 N

and the direction of force is downwards.

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