A 0.375 kg particle slides around a horizontal track. The track has a smooth ver

Question

A 0.375 kg particle slides around a horizontal track. The track has a smooth vertical outer wall forming a circle with a radius of 1.80 m. The particle is given an initial speed of 7.00 m/s. After one revolution, its speed has dropped to 6.00 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.
____?___ J

(b) Calculate the coefficient of kinetic friction.
____?______

(c) What is the total number of revolutions the particle makes before stopping?
___?____ rev

Explanation / Answer

1)

KE = 1/2 m(vi2 - v2f) =0.375 * (49- 36) = 4.9 J

2)

The distance around the track C = 2 * 1.8 = 11.3 m

Work done by friction = Ff * C   so   Ff = 4.9 / 11.3 = 0.431 N (the frictional force)

Ff = m g   and    = 0.431 / (0.375 * 9.8) =0 .12

3)

The total work done in stopping the particle equals the initial KE

The initial KE = 1/2 m vi2 =0.375 * 49 = 18.4

Then    n Ff C = 18.4 J   so n = 18.4 / (0.431 * 11.3) = 3.77 rev

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