A 0.350 kg particle slides arounda horizontal track. The track has a smooth vert

Question

A 0.350 kg particle slides arounda horizontal track. The track has a smooth vertical outer wallforming a circle with a radius of 1.50m. The particle is given an initial speed of 7.00 m/s. After one revolution, its speed hasdropped to 5.00 m/s because offriction with the rough floor of the track. (a) Find the energy transformed from mechanicalto internal in the system as a result of friction.
1 J

(b) Calculate the coefficient of kinetic friction.
2

(c) What is the total number of revolutions the particle makesbefore stopping?
3 rev
A 0.350 kg particle slides arounda horizontal track. The track has a smooth vertical outer wallforming a circle with a radius of 1.50m. The particle is given an initial speed of 7.00 m/s. After one revolution, its speed hasdropped to 5.00 m/s because offriction with the rough floor of the track. (a) Find the energy transformed from mechanicalto internal in the system as a result of friction.
1 J

(b) Calculate the coefficient of kinetic friction.
2

(c) What is the total number of revolutions the particle makesbefore stopping?
3 rev
A 0.350 kg particle slides arounda horizontal track. The track has a smooth vertical outer wallforming a circle with a radius of 1.50m. The particle is given an initial speed of 7.00 m/s. After one revolution, its speed hasdropped to 5.00 m/s because offriction with the rough floor of the track. (a) Find the energy transformed from mechanicalto internal in the system as a result of friction.
1 J

(b) Calculate the coefficient of kinetic friction.
2

(c) What is the total number of revolutions the particle makesbefore stopping?
3 rev
(a) Find the energy transformed from mechanicalto internal in the system as a result of friction.
1 J

(b) Calculate the coefficient of kinetic friction.
2

(c) What is the total number of revolutions the particle makesbefore stopping?
3 rev

Explanation / Answer

(a)    the energy transformed from mechanical tointernal in the system as a result of friction will be    Eint = - K            = - (1 / 2) m (vf2 -vi2)            = - (1 / 2) (0.350 kg) [(5.00 m / s)2 - (7.00 m /s)2]            = ......... J (b)    Eint = f r             =k m g 2 r    k = ......... (c)    suppose we assume that after N number ofrevolutions the object comes to rest and Kf = 0    Eint = - K             =- 0 + Ki             =(1 / 2) m vi2    k m g [N (2 r)]= (1 / 2) m vi2    N = (1 / 2) m vi2/ k m g 2 r        = ........ rev    k m g [N (2 r)]= (1 / 2) m vi2    N = (1 / 2) m vi2/ k m g 2 r        = ........ rev    N = (1 / 2) m vi2/ k m g 2 r        = ........ rev

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