Question

Two floors in a building are separated by 4.1 m. People move between the two floors on a set of stairs. (a) Determine the change in potential energy of a 3.0 kg backpack carried up the stairs. (b) Determine the change in potential energy of a person with weight 650 N that descends the stairs.

Answer 1

Answer:

a) The change in potential energy of a 3.0 kilograms backpack carried up the stairs.

b) The change in potential energy of a persona with weight 650 newtons that descends the stairs is -2665 joules.

Explanation:

Let consider the bottom of the first floor in a building as the zero reference ($z = 0\,m$). The change in potential energy experimented by a particle ($\Delta U_{g}$), measured in joules, is:

$\Delta U_{g} = m\cdot g\cdot (z_{f}-z_{o})$ (1)

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$z_{o}$, $z_{f}$ - Initial and final height with respect to zero reference, measured in meters.

Please notice that $m\cdot g$ is the weight of the particle, measured in newtons.

a) If we know that $m = 3\,kg$, $g = 9.807\,\frac{m}{s^{2}}$, $z_{o} = 0\,m$ and $z_{f} = 4.1\,m$, then the change in potential energy is:

$\Delta U_{g} = (3\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.1\,m-0\,m)$

$\Delta U_{g} = 120.626\,J$

The change in potential energy of a 3.0 kilograms backpack carried up the stairs.

b) If we know that $m\cdot g = 650\,N$, $z_{o} = 4.1\,m$ and $z_{f} = 0\,m$, then the change in potential energy is:

$\Delta U_{g} = (650\,N)\cdot (0\,m-4.1\,m)$

$\Delta U_{g} = -2665\,J$

The change in potential energy of a persona with weight 650 newtons that descends the stairs is -2665 joules.