Question

Problem 6.29 TO of 13 Constants A 72.5-kg hiker starts at an elevation of 1300 m and climbs to the top of a peak 2590 m high Part A What is the hicer's change in potential energy? Express your answer to three significant figures and include the appropriate units. APE= Value Units Submit Request Answer Part 3 What is the minimum work required of the hiker? Express your answer to three significant figures and include the appropriate units.

Answer 1

Ans: According to data given - Mass of hikes M = 72.5 Kg Initial height of hikes Hi= 1300m Final height of hikes Hf= 2590m Partlay: Let all be the change in Patential Then by the formula of potential energy U = MgH AU = M9 AH = Mg lHf-Hi) AU = 72.5 X9:8x (9590-1300) I Au = 79.5 x 9.8 x 1890J DU = 916545 J Hence- Change in Potential Energy: 916545Ans energy of hiker. Postcb): The fosses acting on the hikes as- (1) Mg vestically downward due to earth (2) Ro Contact fosce due to surface below hiker's feet. Nowo - koskdone by Mgr by = -AU [ey the definition of Ally. Woskdone by R W₂ =o[': point of application of A does not move. I Hence- Tatal work done on the hiker by external forces W = W, = WE-916545 J =-40

- lost by Continued But we know hikes is_dided. Why? Because R - pushes the muscles of feet and legs and hence muscles also push the surface below feet in accosdance with Aetion Reaction fosce. So We cam say that energy muscles is equal to work done by hikes. Let WH be the workdone by hiker them - According to energy Conservation- WH + W = 0 WH-916545J = 0 [OR WH = 100/1. Wy = 916545J Hence- Work by Hikes = 916 545 J WH = -AU
Problem is solved using the formula of change in potential energy and basic reasoning.