small object of mass m = 1.40 kg moves in a horizontal circle of radius r = 1.03
ID: 2056006 • Letter: S
Question
small object of mass m = 1.40 kg moves in a horizontal circle of radius r = 1.03 m on a rough table. It is attached to a horizontal string fixed at the center of the circle. The speed of the object is initially 3.79 m/s. After completing one full trip around the circle, the speed of the object is 0.95 m/s. Find the energy dissipated by friction during that one revolution in Joules. Do not enter unit.What is the coefficient of kinetic friction? How many more revolutions will the object be making before coming to rest?Explanation / Answer
energy dissipated =change in kinetic energies=(0.5*1.40)((3.79^2)-(0.95^2))=9.423 J
fictional force =mg*coefficient = 9.423/(distance travelled)=9.423/(2**1.03)
coefficent=9.423/(2**1.03*1.4*9.8)=0.106
kinetic energy should become 0---> enrgy dissipated =0.5*1.40*(3.79^2)=10.054J
coefficeint*(2*3.14*1.03)*no. of revolutions =(10.054)/(mg)
no. of revolutions =(10.054)/((2*3.14*1.03)*1.40*9.8*.106)=1.068
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