Answer to Question #180052 in Classical Mechanics for Jethro

Question

What is the potential energy of 800-kg elevator at the top of a 380 meter tall building?

Answer 1

The potential energy is given by.

$U_{g}=m\;\;g\;h$

Where.

$h=380\;\text{m}$ is the height above the ground.

$g=9.81\;\text{m}/\text{s}^{2}$ is the acceleration of gravity.

$m=800\;\text{Kg}$ is the mass

Evaluating numerically.

$U_{g}=m\;\;g\;h\\ U_{g}=800\;\text{Kg}\times 9.81\;\text{m}/\text{s}^{2}\times 380\;\text{m}\\ U_{g}=2,982,240\;\text{J} \\$

Expressing in scientific notation with three significant figures.

$U_{g}=2.98\times 10^{6}\;\text{J}$

The potential energy is $\displaystyle \color{red}{\boxed{U_{g}=2.98\times 10^{6}\;\text{J}}}$