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An airplane weighing 187,000 lbs is flying parallel to the earth at an altitude of 39,475 ft, when the engine suddenly dies. After the pilot successfully ejects, the plane crashes to the earth. What is the potential energy of the plane in Joules before it starts to fall?

Respuesta :

valeriagonzalez0213

Answer:

[tex]E_{p}=10GJ[/tex]

Explanation:

We apply the formula to calculate the potential energy (Ep):

[tex]E_{p}=m*g*h[/tex]

[tex]W=m*g=187000lbs[/tex]

m: mass

g: acceleration due to gravity

h: height in ft

[tex]h=39475 ft[/tex]

[tex]E_{p}=W*h[/tex] Formula (1)

We replace the data in the formula (1)

[tex]E_{p}=187000*39475[/tex]

[tex]E_{p}=7.3818*10^{9} ft-lb[/tex]

1 ft-lb = 1.3558 J

[tex]E_{p}=7.3818*10^{9}  ft-lb*\frac{1.3558J}{1 ft-lb}[/tex]

1GJ = 1*10^9J

[tex]E_{p} = 10*10^{9} = 10GJ[/tex]

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