A movie stuntman (mass 80.0 kg) stands on a window ledge 5.0 m above the floor (?Fig. P8.81?). Grabbing a rope attached to a chandelier, he swings down to grapple with the movie’s villain (mass 70.0 kg), who is standing directly under the chandelier. (Assume that the stuntman’s center of mass moves downward 5.0 m. He releases the rope just as he reaches the villain.) (a) With what speed do the entwined foes start to slide across the floor? (b) If the coefficient of kinetic friction of their bodies with the floor is µk = 0.250, how far do they slide?
Solution 85P Introduction From the potential energy we can calculate the velocity of the stuntman just before hitting the villain. Now from the momentum conservation, we can calculate the final velocity of them together. Now we will be able to calculate the initial kinetic energy of them combined together. Also they will stop when the work done by friction will be equal to the initial kinetic energy of them combined. From this we can calculate the distance they will slide. Step 1 The mass of the stunt man is m = 80 kg and the mass of the villain is m = 70 kg. Also the initial s v height of the stunt man is h = 5.0 m. Hence the initial potential energy of the stuntman is Now if the velocity of the stuntman just before hitting the villain ss v , then the kinetic energy of the stuntman is Now equating this two energy we have Now if m ic the combined mass of villain and stuntman and v is thcir velocity just after collision, then we have The entwined foes will start sliding with 5.28 m/s speed.
