A \(62-\mathrm{kg}\) woman takes \(6.0 \mathrm{s}\) to run up a flight of stairs. The landing at the top of the stairs is \(5.0 \mathrm{m}\) above her starting place. (a) What is the woman's average power output while she is running? (b) Would that be equal to her average power input-the rate at which chemical energy in food or stored fat is used? Why or why not?

First, we will find the gravitational potential energy gained by the woman as she climbs the stairs. We can find this using the formula: \(PE = m \cdot g \cdot h\) where \(PE\) is the potential energy, \(m\) is her mass, \(g\) is gravitational acceleration (\(9.8 \, m/s^2\)), and \(h\) is the height of the stairs. Plugging in the given values: \(PE = 62 \, kg * 9.8 \, m/s^2 * 5.0 \, m\)

Now we can calculate the gravitational potential energy: \(PE = 62 \, kg * 9.8 \, m/s^2 * 5.0 \, m = 3042 \, J\) (joules)

Now that we have the work done (potential energy gained), we can find the average power output by dividing it by the time taken: \(P_\text{out} = \frac{PE}{t}\) \(P_\text{out} = \frac{3042 \, J}{6.0 \, s}\)

Now we can solve for average power output: \(P_\text{out} = \frac{3042 \, J}{6.0 \, s} = 507 W\)

Now let's discuss the average power input. The woman's body uses chemical energy stored in the form of food or fat. The average power output represents the rate at which this energy is converted into work to climb the stairs. However, not all of the energy stored in the body is used for this purpose - some energy is lost as heat, and some might be used for other bodily activities. Therefore, we can conclude that the average power input will be higher than the average power output found in Step 4. To summarize: - (a) The woman's average power output while running up the stairs is 507 W. - (b) That would not be equal to her average power input, as some of her body's stored energy is lost as heat and used for other purposes.