A movie stuntman (mass 80.0 kg) stands on a window ledge 5.0...

A movie stuntman (mass 80.0 kg) stands on a window ledge 5.0 m above the floor. Grabbing a rope attached to a chandelier, he swings down to grapple with the movie's villain, who is standing directly under the chandelier. (Assume that the center of mass of the stuntman moves downward 5.0 m, which is also the length of the rope. He releases the rope at the bottom, just as he reaches the villain.) a) What is the speed of the stuntman just before he hits the villain? b) What is the tension in the rope just before the stuntman reaches the lowest point? c) If they start to slide over the floor with a speed of 4.8 m/s, what is the mass of the villain? d) As they slide over the rough floor, how much work must be done by friction to bring them to a full stop? e) If the coefficient of kinetic friction of their bodies with the floor is ?k = 0.225, how far do they slide?

Transcript

00:01 Hi, here in this given problem, mass of the stunt man that is given as m1 is equal to 80 .0 kilogram and mass of the villain, the movie, that is m2 is equal to, and this is missing, we have to find it.

00:34 Length of the rope which is being used by the stuntman to come down so the length of the rope will be equal to distance covered by the stuntman in downward direction so we will take it as h and that is 5 .0 meter so now in the first part of the problem to find the speed of the stuntman just before he hits the villain we will use energy conservation and it says kinetic energy of the stuntman half m1 v1 i square at the ground it will be equal to his gravitational potential energy at the height cancelling m1 we get an expression for v1i and that is equal to square root of 2g.

01:45 So this is, this will be given by square root of 2 times of 9 .8 into height that is 5 meter.

01:56 Means square root of 98 meter per second which comes out to be equal to 9 .9 meter per second.

02:04 Answer for the first part of this given problem here then for the second part of the same problem we have to find tension in the rope at the lowermost point so this is the rope this is the stuntman grabbing the rope here so weight of the stentman acting downward vertically downward and considering this circular motion a centrifugal force that will also be acting radially outward mv square by l means h so the tension in the string that will be equal to m g and here it will be m1 mass of the stentment plus m1 v square by h so plugging in all the known values for m1 that is taken out as common into g 9 .8 plus v square will come out to be 98 and h this is 5 so tension is calculated to be equal to 2 ,352 newton which becomes the answer for the second part of this given problem here then in the third part of the problem as they are entangled after collision means they becomes a single mass.

03:45 So using conservation of linear momentum we can say initial momentum of the stuntman m1 into v1i plus that of the villain which is at rest so it will be 0 is equal to final momentum m1 plus m2 into v final velocity and this final velocity is given as 4 .8 meter per second so plugging in all other known values for m m1, this is 8t into v1i that is calculated in the first part of the problem to be 9 .9 is equal to m1 again 8 t plus m2 this is missing we have to find it into final velocity 4 .8.

04:28 So this 8 t plus m2 comes out to be equal to 164 .99.

04:37 So finally mass of the villain that is equal to 164 .99 minus 80...

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