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

Verified Answer :

dexteright02

We have:

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

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
\(TE = \frac{1}{2} m*v^2 + m*g*h\)

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

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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