Calculate Power Output: 20 Press-Ups in 50 Seconds - Physics Problem

In summary, the conversation is about a man who does 20 press-ups in 50 seconds and the question of what his power output is. The conversation discusses different formulas and approaches to finding the answer, including calculating work and power based on the percentage of the man's bodyweight involved in each press-up, estimating the distance of the press-up, and using personal weight and arm length as an estimate. The conversation concludes with the suggestion to not overthink the question and to focus on understanding the concepts of work and power.
  • #1
alias25
197
0
heres this question:
theres a man who does 20 press-ups (ie facing down body straight) in 50 sec what is his power output?

is it 20/50 = 0.4 press-ups per sec of power?

i know power is in watts and its P = Fv = E/t

if i take accleration to be 9.8 i could get v, a= (v-u)/t
initial vel would be 0 ithink but i still can't get force i require mass which I am not given in the Q so should i assume a mass?
or is the answer above correct? and I am thinking too much into it?
 
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  • #2
You'll need to estimate how much work the man does during each press-up. He is lifting some percentage of his bodyweight through some distance.
 
  • #3
isnt he lifting his whole body so 100%of his body weight so force would be 9.8 * (100% of mass) =
i could use S = 1/2 at^2 to find displacement? i tried that and i get a silly answer of 7. summin meters (thats using t as (50/20)/2 and a as 9.8 initial vel as 0)
and say i could get vel by a= (v-u)/t

then mav= mas/t = P
help how do i find some % of bodyweight and some dist.
 
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  • #4
alias25 said:
isnt he lifting his whole body so 100%of his body weight so force would be 9.8 * (100% of mass) =
What matters is how much force he exerts. An example: What's easier, lifting one end of a sofa off the ground or lifting the entire sofa?

Why not check this yourself? Find a bathroom scale. Stand on it to get your weight. Then do a pushup on it, and see what it reads.

i could use S = 1/2 at^2 to find displacement? i tried that and i get a silly answer of 7. summin meters (thats using t as (50/20)/2 and a as 9.8 initial vel as 0)
and say i could get vel by a= (v-u)/t
Those are formulas for freely falling bodies. I don't see anything falling here. The distance? Hint: How long are your arms?
 
  • #5
m*9.8*12.25 = m*9.8*7.66 but the mass cancels and the numbers are silly no one got arms 7 meters long
 
  • #6
my arms are very short lol. so are u suggesting i should assume values? like the average persons weight and arm length?etc
 
  • #7
hang on my arm length is about a third of the length of my body or summin should i be thinking along those lines? ok maybe not a third. ( I've got a new teacher for physics, i don't want him to think I am dumb or something)
 
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  • #8
If you're not sure how far you press yourself up, get a ruler and check. All you need is an estimate. (Did you check what percentage of your weight is involved in a press up? Do it!)

This problem is just an exercise to see if you understand the ideas of work and power. All you will count, I presume, is the positive work done as the man pushes himself up. You can use your own weight/force/distance as an estimate, or you can express it in terms of the unknown man's weight and arm length. (But don't start talking about falling objects! :wink: )
 
  • #9
ok :blushing: thanks i think i did look too much into the question.
 

FAQ: Calculate Power Output: 20 Press-Ups in 50 Seconds - Physics Problem

What is power output in physics?

Power output in physics refers to the amount of work done or energy transferred per unit time. It is measured in watts (W) and is a measure of how fast energy is being used or produced.

How is power output calculated?

Power output is calculated by dividing the work done or energy transferred by the time it takes to do the work or transfer the energy. The formula for power output is: Power (W) = Work (J) / Time (s).

What factors affect power output?

The factors that affect power output include the amount of work done or energy transferred, the time it takes to do the work or transfer the energy, and the efficiency of the system or device. Other factors such as temperature, friction, and resistance may also affect power output.

Can power output be increased?

Yes, power output can be increased by increasing the amount of work done or energy transferred, decreasing the time it takes to do the work or transfer the energy, or by improving the efficiency of the system or device.

What are some real-life examples of power output?

Some real-life examples of power output include the power of a car engine, the power of a light bulb, and the power of a human in terms of physical work. Power output is also used in the measurement of electricity, such as the power output of a household appliance or power plant.

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