As shown in the figure below, a green bead of mass 50 g slides along a straight

Question

As shown in the figure below, a green bead of mass 50 g slides along a straight wire. The length of the wire from point to pointis 0.700 m, and point is 0.400 m higher than point D. A constant friction force of magnitude 0.0050 N acts on the bead. (a) If the bead is released from rest at point , what is its speed at point B m/s (b) A red bead of mass 50 g slides along a curved wire, subject to a friction force with the same constant magnitude as that on the green bead. I the green and red beads are released simultaneously from rest at point , which bead reaches point speed? with a higher the green bead the red bead both beads arrive with equal speed Explain

Explanation / Answer

Since B is 0.4m below A, the bead has lost 'mgh' potential energy and picked up that much KE. But then in sliding down the wire it has lost F x d energy to friction. Let's sum these...

dU = mgh - Fd
dU = 0.050kg (9.81m/s^2) 0.400m - 0.0050N * 0.700m
dU = 0.1925 J

dU = KE = .5 m v^2
v^2 = 2 KE / m
v = sqrt(2 KE / m)
v = sqrt(2 (0.1925J) / .030kg)
v = 2.774 m/s

B)

The green bead will reach point B at a higher speed because it traveled a shorter distance, and therefore there was a shorter distance for the frictional force to act

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