Question

In: Physics

Robot No. 5 (mass 273kg) and Prisoner No. 6 (mass 66kg) stand on a frictionless ice...

Robot No. 5 (mass 273kg) and Prisoner No. 6 (mass 66kg) stand on a frictionless ice rink, both holding a rope. The robot is at x = −3m and the prisoner is at x = 17m. The robot starts winding up the rope. Considering the two of them plus the rope as a system, there are no external horizontal forces on either of them. When the rope is fully wound, where do No. 5 and No. 6 meet?

Solutions

Expert Solution

here robot no-5 is standing at , x= -3 m ,

prisoner-6 is standing at , x= 17 m ,

mass of robot no-5 = 273 kg

mass of prisoner-6 = 66 kg

so here we can draw a simple diagram ,

for two mass system , we have to find center of the mass of the system where they both meet ,

we have formula for finding center of mass of two system = m1 x1 + m2 x2 / m1 + m2

here for robots and prisoner system,

Xcm = mrobot x1 + mprisoner x2 / mrobot​ + mprisoner

here mrobot = 273 kg

  mprisoner = 66 kg

here X1 = here robot no-5 is standing at , x= -3 m , X1 = -3 m

here X2 =  prisoner-6 is standing at , x= 17 m ,   X2 = 17 m

so here center of the mass of system where they both meet ,

Xcm = mrobot x1 + mprisoner x2 / mrobot​ + mprisoner

= 273* (-3) + 66 (17) / (273+66)

Xcm =303 / 339

Xcm = 0.89 m

so they both are meet at Xcm = 0.89 m

here when robot-5 and prisoner-6 meet's at  Xcm = 0.89 m , then rope is fully wound.


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