A 55.0-kilogram woman in a gym does
50 deep knee bends in 3.00 minutes. In each knee bend, her center of
mass is lowered and raised by 0.400 meters. She does work in both
directions. Assume her efficiency is 20
percent. Calculate the energy used to do all
50 repetitions in units of kilojoules. What is her average power
consumption in watts?
In this two-part exercise, we want
to solve for the energy — we can call it capital 𝐸 — and the average power
consumption — we can call it capital 𝑃 — involved in this woman exercising. Let’s start with a diagram. In this scenario, a woman starts
out upright and in doing a deep knee bend lowers her center of gravity at distance ℎ
given as 0.400 meters. She then stands back up and raises
her center of gravity that same vertical distance. We’re told the woman does 50 of
these deep knee bends in a time 𝑡 of 3.00 minutes and has an efficiency we’ve
called lowercase 𝑒 of 20 percent or 0.20.
We wanna start by solving for the
total energy required for her to go through 50 of these motions. We can say that the total energy
output by the woman is equal to the total work she does divided by the efficiency of
that work against gravity. Recalling that work is equal to
force times distance, we can write that the total work 𝑤 is equal to the force of
gravity on the woman, 𝑚 times 𝑔, multiplied by the distance she moves through
gravity, ℎ, all multiplied by two times the number of repetitions she does. We’ll have that factor of two
because she moves down and then up.
Treating the acceleration due to
gravity as exactly 9.8 meters per second squared, we see we’ve been given the
efficiency 𝑒, mass 𝑚, we know 𝑔, given ℎ, and we know capital 𝑁. So we’re ready to plug in and solve
for 𝐸. Plugging in the values for the
efficiency 𝑒, mass, 𝑔, ℎ, and two times 𝑁, when we calculate the total energy
used 𝐸, we find it’s equal to 108 kilojoules. That’s the total amount of energy
the woman would need to supply to do all these exercises.
Next, we want to solve for the
power output of the woman over the course of these exercises. Recalling that power is equal to
energy per unit time, we can write that 𝑃 is equal to 𝐸, the result we solved for
in part one, divided by 𝑡, the time value given as 3.00 minutes. When we plug in our two values,
we’re careful to convert our time into units of seconds. That’s because power is given in
units of joules per second or watts. Calculating this fraction, to three
significant figures, it’s 599 watts. That’s the power the woman would
need to supply to do these exercises in this amount of time.