oo T-Mobile 10:34 PM @ 33% spock.physast.uga.edu Course Contents » »Problem Set

Question

oo T-Mobile 10:34 PM @ 33% spock.physast.uga.edu Course Contents » »Problem Set 8» (1) 'Timer LI Notes Evaluate tw/ Feedback rint Info A 0.444 kg particle slides around a horizontal track The track has a smooth, vertical outer wall forming a circle with a radius of 1.70 m. The particle is given an initial speed of 8.95 m/s. After one revolution, its s has dropped to 6.50 m/s because of friction with the rough floor of the track. Calculate the energy loss due to friction in one revolution Submit Answer Tries 0/8 Calculate the coefficient of kinetic friction Suemi lower Ties olt revolutions the partide What is the total number of revolutions the particle makes before stopping? Do not enter units. Subnmit Answer Tries 0/s Post Discussion Send Feedback

Explanation / Answer

First Part:

Here, 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 =6.50 m/s
vi = initial velocity =8.95 m/s
m = mass = 0.444 kg

So energy loss due to friction in one revolution = change in kinetic energy

= 0.5*0.444*(8.95^2 - 6.5^2) = 8.40 J

Second Part:

s = distance= circumference = 2*pi*r= 10.68 m
vf^2 -vi^2 =2as
a= (vf^2 -vi^2)/(2s)= -1.77 m/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 = 1.77/9.81 = 0.18

Third Part:
Again use the expression -

vf^2 -vi^2 =2as
this time final velocity is zero vf= 0
vi= 8.95 m/s
a= - 1.77 m/s^2
s= vi^2/(2a) = 22.63 m

So, no. of revolutions before stopping = s/(circumference) = 2.11

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.