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a 0.112 kg apple on a tree has a PE of 4.00 J. How high is it above the ground?

 Jul 1, 2019
 #1
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What is the "PE" formula in your textbook?

 Jul 1, 2019
 #2
avatar+8731 
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a 0.112 kg apple on a tree has a PE of 4.00 J. How high is it above the ground?

 

Hello Guest!

 

\(F=\frac{G\cdot m\cdot M}{r^2}\\ r=\sqrt{\frac{G\cdot m\cdot M}{F}}\)

 

F = PE = 4J

m = 0.112 kg

M = \(5.97\cdot 10^{24}\)kg

G = \(\frac{6.672\cdot 10^{-11}m^3}{kg\cdot s^2}\)

 

\(r=\sqrt{\frac{\frac{6.672\cdot 10^{-11}m^3}{kg\cdot s^2}\cdot 0.112kg\cdot 5.97\cdot 10^{24}kg}{4\frac{kg\cdot m}{s^2}}}\)

\(r=\sqrt{\frac{m^3\cdot kg\cdot kg\cdot s^2}{kg\cdot s^2\cdot kg\cdot m}\times \frac{6.672\cdot 10^{-11}\cdot 0.112\cdot 5.97\cdot 10^{24}}{4}}\)

\(r=\sqrt{9152\cdot 10^{13}\ m^2}\\ r=3339598.05965\ m\)

\(high=3339598.05965\ m-6378 m (Earth's\ radius, geocentric \ mean)\\ \color{blue} high=3333220\ m\)

 

The tree stands on the south pole and is about 3333221m high.

 

Sorry, I read 4 N instead of 4 J.

 

\(PE= m\times g\times h\\ h=\frac{PE}{m\times g}\\ h=\frac{4J\times s^2}{0.112kg\times 9,81m}\times\frac{kg\times m^2}{J\times s^2}\\ \color{blue}h=3.641m\)

laugh  !

 Jul 2, 2019
edited by asinus  Jul 2, 2019
edited by asinus  Jul 2, 2019
 #4
avatar+1821 
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Asinus, Are you nibbling on pot-laced candy?   

...Save some for me  laugh

 

 

GA

GingerAle  Jul 2, 2019
 #6
avatar+8731 
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GA, please no flattery.
Tell me what is wrong!

laugh  !

asinus  Jul 2, 2019
 #3
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PE = m x G x h, where m=mass, G =Gravitational constant, h =Height above the ground.

4 = 0.112 x 9.81 x h

4 = 1.09872 x h 

h = 4 / 1.09872

h = 3.64 meters above the ground.

 Jul 2, 2019
 #5
avatar+1821 
0

Congrats Mr. BB, your answer is correct; however, the identifier (G) gravitational constant in this equation should be (g) acceleration of gravity (earth average).

 

 

GA

GingerAle  Jul 2, 2019
edited by GingerAle  Jul 2, 2019

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