Consider the system shown in the figure (Figure 1) . The rope and pulley have ne
ID: 1902087 • Letter: C
Question
Consider the system shown in the figure (Figure 1) . The rope and pulley have negligible mass, and the pulley is frictionless. Initially the 6.00-{ m kg} block is moving downward and the 8.00-{ m kg} block is moving to the right, both with a speed of 0.300m/s . The blocks come to rest after moving 9.00m .no title provided
Use the work-energy theorem to calculate the coefficient of kinetic friction between the 8.00-{ m kg} block and the tabletop.
Looking for the coefficient of friction, not speed... Thanks
Explanation / Answer
initial energy of system=
6kg block= mgh+0.5mv^2= 6[9.8*9+0.5*0.3^2]=529.47
8kg block= 0.5mv^2= 0.5*8*0.3^2=0.36
final energy of system=0
we here assume ground level to be 9m below the 6kg mass
work = change in energy
w=529.47+0.36=529.72
this work is done by friction force mg, this is = work done
mg*s=w
=529.72/[9.8*8*9]=0.75