harness and friction brake to a rope suspended from point P at the top of the building.Figure 8 shows a man cleaning windows on a tall building. He is attached by a The rope has a mass of 18 kg and an unstretched length of 295 m. The rope obeys Hooke's law. The ground is 309 m below P. The total mass of the man, including his equipment, is 98 kg.

Explain why the maximum tension in the rope is approximately 1140 N when the man is sliding down the rope at a constant speed of 1.5 ms¹. When the man arrives at a dirty window, he increases F. An impulse acts on him and he decelerates from 1.5 m s¹ to rest. Show that the magnitude of this impulse is approximately 150 N s. Figure 9 shows the variation of F with time. The graph starts before the impulse begins and extends for a short time after the impulse ends.

The man reaches the bottom of the rope and stops. He is now suspended in equilibrium above the ground. ]Show that the average tension in the rope is now approximately 1050 N.[2 5Calculate the distance above the ground of the bottom end of the rope. cross-sectional area of rope = 3.14 cm² Young modulus of the rope material = 3.51 x108 Pa