Use the worked example above to help you solve this problem. A 59.0 kg skier is
ID: 1289532 • Letter: U
Question
Use the worked example above to help you solve this problem. A 59.0 kg skier is at the top of a slope, as shown in the figure. At the initial point circled A, she is 9.0 m vertically above point circled B. (a) Setting the zero level for gravitational potential energy at circled B, find the gravitational potential energy of this system when the skier is at circled A and then at circled B. Finally, find the change in potential energy of the skier-Earth system as the skier goes from point circled A to point circled B. PEi = Correct: Your answer is correct. J PEf = Correct: Your answer is correct. J PE = Correct: Your answer is correct. J (b) Repeat this problem with the zero-level at point circled A. PEi = Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. J PEf = J PE = J (c) Repeat again, with the zero level 2.00 m higher than point circled B. PEi = J PEf = J PE = J EXERCISE HINTS: GETTING STARTED | I'M STUCK! Use the values from PRACTICE IT to help you work this exercise. If the zero level for the gravitational potential energy is selected to be midway down the slope, 4.50 m above point circled B, find the initial potential energy, the final potential energy, and the change in potential energy as the skier goes from point circled A to circled B in the figure. initial kJ final kJ change kJ
Explanation / Answer
mass of the skier m = 53 kg
vertical height h = 9 m
a)
initial potential energy of the skier is
(P.E)i = mgh
= (53 kg)(9.8 m/s2)(9 m)
= 4674.6 J
= 4.67*103 J
at bottom (height h = 0 m) ,
final potential energy is
(P.E)f = mgh
= mg(0)
= 0 J
difference between in potential energy between A and B is
(P.E)f - (P.E)i = 0 J - 4.674*103 J
= - 4.67*103 J
..............................................................
b)
potential energy of the skier is
(P.E)i = mgh
= 0 J
potential energy (P.E)f = mgh
= (53 kg)(9.8 m/s2)(-9 m)
= - 4.67*103 J
difference between in potential energy is
(P.E)f - (P.E)i = - 4.67*103 J - 0 J
= - 4.67*103 J
.............................................................................
c)
potential at point A :
(P.E)i = mgh
= (53 kg)(9.8 m/s2)(9 m - 2 m)
= 3635.8 J
= 3.63*103 J
potential at point B :
(P.E)f= mgh
= (53 kg)(9.8 m/s2)(- 2 m)
= -1038.8 J
= -1.04*103 J
change in potential energy is
(P.E)f - (P.E)i = - 1.04*103 J - 3.63*103 J
= - 4.67*103 J
.........................................................................................
.........................................................................................
initial potential energy is
(P.E)i = mgh
= (53 kg)(9.8 m/s2)(4.5 m)
= 2337.3 J
= 2.337*103 J
= 2.337 kJ
final potential energy is
(P.E)f = mgh
= (53 kg)(9.8 m/s2)(-4.5 m)
= -2337.3 J
= -2.337*103 J
= -2.337 kJ
change in potential energy is
(P.E)f - (P.E)i = - 2.337*103 J - 2.337*103 J
= - 4.67*103 J
= - 4.67 kJ
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