Problem 7-50a: A particle of mass m =2.00kg moves in a horizontal circle of radi
ID: 1495252 • Letter: P
Question
Problem 7-50a: A particle of mass m=2.00kg moves in a horizontal circle of radius r=1.31m on a rough table. It is attached to a horizontal string fixed at the center of the circle. The speed of the particle is initially 3.65m/s. After completing one full trip around the circle, the speed of the particle is 1.46m/s. Find the energy dissipated by friction during that one revolution in Joules.
(in J)
E: 13.99
Problem 7-50b: A particle of mass m=2.00kg moves in a horizontal circle of radius r=1.31m on a rough table. It is attached to a horizontal string fixed at the center of the circle. The speed of the particle is initially 3.65m/s. After completing one full trip around the circle, the speed of the particle is 1.46 m/s. Find the coefficient of kinetic friction.
Problem 7-50c: A particle of mass m=2.00kg moves in a horizontal circle of radius r=1.31m on a rough table. It is attached to a horizontal string fixed at the center of the circle. The speed of the particle is initially 3.65m/s. After completing one full trip around the circle, the speed of the particle is 1.46 m/s. How many more revolutions will the particle make before coming to rest? Enter your answer as a number using the decimal place.
A: 5.73 B: 7.16 C: 8.95 D: 11.19E: 13.99
Explanation / Answer
work done = change in kinetic energy
W = 0.5*m*(v^2-u^2)
W = 0.5*2*(1.46^2-3.65^2)
W = -11.19 J
dissipated = 11.19 J
option D
+++++++++++++++++++
Work done = uk*m*g*s = uk*m*g*2*pi*r
11.19 = uk*2*9.8*2*pi*1.31
uk = 0.069
+++++++++++
W = change in KE
KEi = 0.5*m*u^2
KEf = 0
W = uk*m*g*n*2*pi*r
n = number of revolutions
0.069*2*9.8*n*2*pi*1.31 = 0.5*2*1.46^2
n = 1.9*10^-1
option B