green bead mass 25 g slides along straight wire length of the wire from A to B is 0.600 m, and point A is 0.200 m higher than point B . A constant friction force of magnitude 0.0250 N acts on the bead

Question

(a) If the bead is released from rest at point A , what is its sp eed at point B ?

(b) A red bead of mass 25 g slides along a curved wire, subject to a friction force with the same constant magnitude as that on the green bead. If the green and red beads are released simultaneously from rest at point A , which bead reaches point B with a higher speed? Explain.

Answer 1

Let

m = 25 g = 0.025 kg be the mass of both beads,

s = 0.600 m be the length of the streight wire,

h = 0.200 m be the height of A above B,

F = 0.0250 N be the force of friction,

v (to be computed) be the speed of the green bead at point B,

g = 9.81 m/s² be gravitational acceleration.

(a)

The potential energy, mgh, at A gets converted to kinetic energy, mv²/2, at B

plus the work of friction, Fs.

This conservation of energy statement is expressed by the equation

mgh = mv²/2 + Fs

So,

v = √2√(gh - Fs/m)

Substituting actual numbers

v = √2√(9.81×0.2 - 0.025×0.6/0.025) = 1.65 m/s

(b)

The speed of the read bead is calculated the same way; the only difference is that it travels along

a longer wire than s.

Therefore friction will perform more work.

Therefore its speed at B will be lower than the speed of the green bead.