A prisoner in jail decides to escape by sliding to freedom down a rope provided by an accomplice. He attaches the top end of the rope to a hook outside his window; the bottom end of the rope hangs clear of the ground. The rope has a mass of 10 kg, and the prisoner has a mass of 70 kg. The hook can stand a pull of 600 N without giving way. If the prisoner's window is 15 m above the ground, what is the least velocity with which he can reach the ground, starting from rest at the top end of the rope? My solution: For the hook I calculate: M-mass rope 10 kg m-mass prisoner 70 kg F –force max. on the hook 600 N h-height 15 m (1) F>T+Mg (2) ma=mg-T From 1 T < F-M and from 2 mg-ma<F-Mg Ma>(m+M)*g-F a>((m+M)*g-F)/M Now v=SQRT(2*a*h) I can find h. But I am not sure for two free body diagram : (1) and (2). Is (1) probably F>T+Mg+mg because prisoner is on the rope or in (2) there is influence of mass of rope or not on the prisoner? Please help how to think at this problems.