A 0.350 kg particle slides around a horizontal track. The track has a smooth ver

Question

A 0.350 kg particle slides around a horizontal track. The track has a smooth vertical outer wall forming a circle with a radius of 1.50 m. The particle is given an initial speed of 7.00 m/s. After one revolution, its speed has dropped to 5.50 m/s because of friction with the rough floor of the track. (a) Find the energy transformed from mechanical to internal in the system as a result of friction. Your response differs from the correct answer by more than 10%. Double check your calculations. J (b) Calculate the coefficient of kinetic friction. (c) What is the total number of revolutions the particle makes before stopping? rev

Explanation / Answer

(a) The energy transformation is simply the difference in kinetic energy:
E = (1/2)(0.35)(72) - (1/2)(0.35)(5.52) = (0.35/2)(72 - 5.52) 3.28 J

(b) This change in kinetic energy is due to the (negative) work done by friction. Work is simply the (frictional) force times the distance over which the force acts, in this case, the circumference of the circle:

W = f x = E

f(2r) = E

f = E/2r

The frictional force is the coefficient times the normal force, which in this case is just the weight:

f = n = w = m g = E/2r

= E/2mgr = 3.28/(2)(0.35)(9.81)(1.5) 0.101

(c) We already have that W = KE

The work done by friction in N revolutions is 2Nrf = 2Nrmg

2Nrmg = mv2/2

Solve for N

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