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Calculate the total energy of 2.0 kg object moving horizontally at 10 m/s 50 meters above the surface

Respuesta :

dexteright02
We have:

Total Energy: KE + GPE
KE (Kinetic Energy) = [tex] \frac{1}{2} m*v^2[/tex]
GPE (Gravitational Potential Energy) = [tex]m*g*h[/tex]

Data:
m (mass) = 2.0 Kg
v (speed) = 10 m/s
h (height) = 50 m
Use: g (gravity) = 10 m/s²

Formula:

Total Energy: KE + GPE
[tex]TE = \frac{1}{2} m*v^2 + m*g*h[/tex]

Solving:
[tex]TE = \frac{1}{2} m*v^2 + m*g*h[/tex]
[tex]TE = \frac{1}{2} *2.0*10^2 + 2.0*10*50[/tex]
[tex]TE = \frac{2.0*100}{2} + 1000[/tex]
[tex]TE = \frac{200}{2} + 1000[/tex]
[tex]TE = 100 + 1000[/tex]
[tex]\boxed{\boxed{TE = 1100\:Joule}}\end{array}}\qquad\quad\checkmark[/tex]

Answer:

The total energy is 1080 J

Explanation:

KE = 1/2 . m . v^2 = 0.5 . 2 . 10^2  = 100J

Height energy ( relative to the surface)  = m . g . h  = 2 . 9.8 . 50  = 980 J

See: https://answers.yahoo.com/question/index?qid=20110215125659AAT5Q0M

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