Video: Work, Energy, and Power in Humans

A 55.0 kg woman in a gym does 50 deep knee bends in 3.00 min. In each knee bend, her center of mass is lowered and raised by 0.400 m. (She does work in both directions.) Assume her efficiency is 20%. Calculate the energy used to do all 50 repetitions in units of kilojoules. What is her average power consumption in watts?

03:06

Video Transcript

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 โ€” we can call it capital ๐ธ โ€” and the average power consumption โ€” we can call it capital ๐‘ƒ โ€” involved in this woman exercising. Letโ€™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 โ„Ž given as 0.400 meters. She then stands back up and raises her center of gravity that same vertical distance. Weโ€™re told the woman does 50 of these deep knee bends in a time ๐‘ก of 3.00 minutes and has an efficiency weโ€™ve called lowercase ๐‘’ 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 ๐‘ค is equal to the force of gravity on the woman, ๐‘š times ๐‘”, multiplied by the distance she moves through gravity, โ„Ž, all multiplied by two times the number of repetitions she does. Weโ€™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 weโ€™ve been given the efficiency ๐‘’, mass ๐‘š, we know ๐‘”, given โ„Ž, and we know capital ๐‘. So weโ€™re ready to plug in and solve for ๐ธ. Plugging in the values for the efficiency ๐‘’, mass, ๐‘”, โ„Ž, and two times ๐‘, when we calculate the total energy used ๐ธ, we find itโ€™s equal to 108 kilojoules. Thatโ€™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 ๐‘ƒ is equal to ๐ธ, the result we solved for in part one, divided by ๐‘ก, the time value given as 3.00 minutes. When we plug in our two values, weโ€™re careful to convert our time into units of seconds. Thatโ€™s because power is given in units of joules per second or watts. Calculating this fraction, to three significant figures, itโ€™s 599 watts. Thatโ€™s the power the woman would need to supply to do these exercises in this amount of time.

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