5. Now a 58.3 kg person climbs the ladder to a point 4.20 m off the floor a. How

Question

5. Now a 58.3 kg person climbs the ladder to a point 4.20 m off the floor a. How much work did he do? b. Taking the ground as a reference (Ep 0) how much gravitational potential energy does this person have at the top of his climb? is person falls, what will be his kinetic energy when he hits the ground? c. If thi ET height E Ep top 3.60 m haltway 1/3 up Fill in the table to the right, using the bottom answers from a, b, and c to help you d. e. Use the formula for kinetic energy to calculate the speed of the man as he hits the ground. problem, we assumed that mechanical energy was conserved. In fact, mechanical energy is we perceive this loss of energy (what does never completely conserved and it becomey? is lost from the system. How do

Explanation / Answer

1 Work doe = f.s = mgs = 58.3*9.8*4.20 = 2399.63 Joule

2 gravitational potential Enegry = Mgh = 58.3*3.9*4.20 = 2399.63 Joule

3 When he will hit the ground all potential energy will become Kinetic Energy, Which is equal to 2399.63 Joule

4 top Ep = 2399.63 Joule , Ek =0, Et = 2399.63 Joule

halfway potential energy will be half and half will become kinetic energy

halfway Ep = 1199.82 Joule, Ek=1199.82 Joule, Et=2399.63 Joule

1/3 up

potential energy will be 1/3 and rest will become kinetic energy

Ep = 799.88 Joule, Ek = 15.99.75 Joule , Et = 2399.63 Joule

bottom

Ep = 0, Ek=2399.63 Joule, Et=2399.63 Joule

e

0.5*mv*v=2399.63

v = 9.07 m/s

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