A 0.400 kg particle slides around a horizontal track. The track has a smooth ver
ID: 1479265 • Letter: A
Question
A 0.400 kg particle slides around a horizontal track. The track has a smooth vertical outer wall forming a circle with a radius of 1.20 m. The particle is given an initial speed of 5.00 m/s. After one revolution, its speed has dropped to 5.00 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.
(b) Calculate the coefficient of kinetic friction.
(c) What is the total number of revolutions the particle makes before stopping?
Explanation / Answer
(a) Use Conservation of Energy, KE_i= KE_f + friction
KE= kinetic energy= 0.5mv^2
change in kinetic energy = 0.5m(vi^2 -vf^2)
vf = final velocity =5.00 m/s
vi = initial velocity =5.00 m/s
m = mass = 0.425 kg
change in kinetic energy, amount of energy transformed from mechanical to internal in the system = 0 J
(b)s = distance= circumference = 2*pi*r= 9.424 m
vf^2 -vi^2 =2as
a= (vf^2 -vi^2)/(2s)= 0m/s^2
sum the forces in the vertical direction
N -mg= 0
N =mg
g= 9.81 m/s^2
sum the forces in the horizontal direction
F- f =ma
f= uk(N)
uk =coefficient of kinetic friction
uk(mg)=ma
uk= a/g =0
(c) infinite (because change in velocity is zero)
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