Question

QUESTION 10

Tony Hawk is skating through a half-pipe with a radius of 4 m. Initially he is in a crouched position with his center of mass 0.5 m above the surface of the half-pipe. He starts from rest with his center of mass a height of 3 m above the base of the half-pipe and experiences no loss of energy from friction as he moves. At the bottom of the half-pipe he stands up and lifts his arms into the air, thereby raising his center of mass to 1.0 m. He then continues up the other side of the half-pipe. Tony had to perform work to lift his arms while skating. How much work (in J) did he do if his mass is 78 kg? You may model Tony as a point particle.

Answer 1

Solution: Let Tony starts from point S where his center of mass is at the height OL 3.5 m from the base of the half pipe. The initial potential energy of the Tony with which he starts from point S is Us mg 78x9.8 x3.5J,m 78 kg and the kinetic energy at point S is K 0. Hence Tony's total mechanical energy of at point S is Ug +K 78x9.8x3.5 J At point B his center of mass is at point C is at height CB- 1 m where he raises his hands Therefore his potential energy is U 78x9.8x1.0 J. Now when he reaches at the other end of the half pipe his height from the base of the half pipe is 4m and velocity is zero (no kinetic energy). Therefore, Tony's potential energy at finishing end O' isUo 78x9.8x 4.0 J. There is no any kind of energy loss. Therefore, Tony's total mechanical energy at O' is UoKo 78x9.8 x4.0 J. The total energy at O' is greater than the total energy at S Therefore, the work done W by Tony is equal to the change in total energy Therefore, W= 78x 9.8 x 4.0J -78x9.8 x 3.5 J =78x9.8 x 0.5 J =382.2J O NA-0.5m, AM-3m NM NA+AM= 0.5m +3m 3.5m A CB=1m C 3m 4m м L B