College Physics by Openstax Chapter 7 Problem 2

A 75.0-kg person climbs stairs, gaining 2.50 meters in height. Find the work done to accomplish this task. (Neglect friction in your calculations.)

Solution:

Work done against gravity in lifting an object becomes potential energy of the object-Earth system. The change in gravitational potential energy is $\Delta PE_{g} = mgh$ , with $h$ being the increase in height and $g$ the acceleration due to gravity.

W=mgh

We are given the following values: $m=75.0\ \text{kg}$ , $g=9.80\ \text{m/s}^2$ , and $h=2.50\ \text{m}$ .

Substitute the given in the formula.

\begin{align*} W & = mgh \\ W & = \left( 75.0\ \text{kg} \right)\left( 9.80\ \text{m/s}^2 \right)\left( 2.50\ \text{m} \right)\\ W & = 1837.5\ \text{Nm} \\ W & = 1837.5\ \text{J} \\ W & = 1.84 \times 10 ^{3} \ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The work done is about $1.84 \times 10 ^ {3}\ \text{Joules}$ .