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2.A 60-kg student does 60 push-ups in 40 s. With each push-up,the student must lift an average of70% of the body mass a height of 40 cm off the floor. Assuming twos...

Question

2.A 60-kg student does 60 push-ups in 40 s. With each push-up,the student must lift an average of70% of the body mass a height of 40 cm off the floor. Assuming twosignificant digits, calculate thefollowing:(a) the work the student does against the force of gravity for eachpush-up, assuming work is doneonly when the student pushes up(b) the total work done against the force of gravity in 40 s(c) the power achieved for this period.

2.A 60-kg student does 60 push-ups in 40 s. With each push-up, the student must lift an average of 70% of the body mass a height of 40 cm off the floor. Assuming two significant digits, calculate the following: (a) the work the student does against the force of gravity for each push-up, assuming work is done only when the student pushes up (b) the total work done against the force of gravity in 40 s (c) the power achieved for this period.



Answers

(a) What force should the woman in Figure 9.46 exert on the floor with each hand to do a push-up? Assume that she moves up at a constant speed. (b) The triceps muscle at the back of her upper arm has an effective lever arm of $1.75 \mathrm{cm},$ and she exerts force on the floor at a horizontal distance of
20.0 $\mathrm{cm}$ from the elbow joint. Calculate the magnitude of the force in each triceps muscle, and compare it to her weight. (c) How much work does she do if her center of mass rises 0.240 $\mathrm{m} ?(\mathrm{d})$ What is her useful power output if she does 25 pushups in one minute?

Oh, welcome to a new problem. So this time we have a man and his son. So they're doing what we call chin ups. Welcome lifting, lifting their bodies up all the way up from the ground. So they're going up and down and then, you know, obviously his son has a shorter hands, maybe have his hands. And so what's happening here is that they have to lift themselves. The mind has to lift uh, himself a distance off zero point four zero meters and so want to find out You know how much work is done in You know, how much work is done? Park il o gram in lifting in, lifting himself up. And you could remember that the mind has a weight. So if he left himself up from this position up until that position walk, Dunne is going to be MGH. But we're interested in finding out, you know, the the work part kilogram. So how much work is done? Part kilogram? Because walk over muss. And that just means that you'll have to divide both sides by him. These to cancel out. So the work over the mass, which is the walk for kilogram his G H G is nine point eight meters per second squared and then the height is lifting himself his zero point four zero meters. And it means that the money three point nine two jewels kilogram to lift himself have a distance of zero point four zero meters in terms ofthe percentages, you know, assumed that and the money he's using up, you know, his muscles are using up seventeen jewels, um off work park, you'LL grow off his Marcel's remember, he's using his muscles to lift himself up. So this is some seventy jewels off walk part kilogram, and you want to compare this to how much work he's doing for killed home. So we want to see what's the this is for? Be part A. This was the answer to what a answered, but is that the muscles are involved in the chin up, so the muscles are using seventy Jules Kilogram and that barely lift him up a distance off zero point four zero meters. So what percentage off his body's muss do this Marcel's constitute? So, um, of one of two, the percent this percentage three point nine two Jules kilogram and that's the work done by the month kilogram body Mass and then compare that to the work done by the Marcel's seven jewels for children. And if you simplify that, no, it's going to help us get the percentage of muscles which happened to me five point six percent. If you are simplify that problem we haven't talked about his son was seem that the certain muscles involved in generating about seventy jewels of what the Pilgrim. And so you know what's percentage off the body's weight constitute thes Marcel's for his son. I'm going to do that in the second page. We have that his hands are short compared to his father. This is the sun, and that means he's gonna lift himself half the distance, which happens to be a zero point two zero meters. So the work pug kilogram that he does he's still going to be md hate because, remember, uh uh, watching the World Park Il aground, the work itself is mth. The walk is doing to lift himself up because he's working against his way on DH to get the walk killed. Mom, we have to divide by the mass s o. His mass happens to be so we don't know what his masses, but the acceleration due to gravity is nine point eight meters per second squared, then multiply by the height that his moving, which is your upon two meters. Remember these to cancel out, and that gives us two point eight percent, which is pretty much, um, not two point eight percent. This one gives us one point nine six Jules kilogram. And if you compare that to the amount off, um, energy expended by his dad a kilogram, it's It's almost half of that, you know? So one point nine six Jules killer from he's muscles expend. So he's muscles. The sun's muscles expand seventy or rather seventy Jules Kilogram. So he's muscles expend seventy Jules Kilogram. And if you compare that to the amount off work he does kilogram on his body mass, it's going to be one point nine six jewels for kill Graham over seventeen jewels for children. Excuse me. And that gives us a two point eight percent in terms ofthe his, um, muscle percentage. Okay, so this is this is the body mass or percentage off his body mass for his muscles. His muscles constitute two point eight percent if you go back. His dad's vassals constitute five point six percent on the other question than asking is, you know, if you compare the DOD and the sun, Um, so who's I was gonna use more jewels for the children. So the sun is on ly using one point nine six, Jules killed Graham to lift himself up. So the sun, our users one point nine six Jules per kilogram, which is less is less done. The fathers three point nine two Jules Kilogram Therefore, to conclude a based on logic that it's easier for the sun to lift himself on chine ups. Done the father. Okay, so this is the last part of the problem. Thanks again for watching the videos. I hope you enjoyed it. Feel free to ask questions and this is a follow up or energy. You know, a son in the father comparing themselves, lifting themselves up the muscles I involved. You want to find out what's thie? Energy? Popular war, more energy. Pipe by unit muss off Bali Ma six expanded so three point nine two jewels for the father. And for the sign it's one four nine six jewels for kilogram the father three point into Jules per killed one. And if you compare the son and the father, you can see that the sun can, you know. Why is it the children can do more chin ups than adults? And you can see that's the reason why Hope you enjoyed the video. Thanks. And see you next time. Okay, bye.

Hi. In the given problem, there is a box having a mass Of M is equal to 30 kg. It is being pulled by two students. Student A is applying a force on it. F. A. Is equal to 30 newton, and this force is constant and student B is applying a force on it. F. B. Which is initially Having a magnet Euro 14 Newton and then gradually it starts reducing. No. For this problem, we use the concept that work done is found by the area off force versus displacement occurred. Now the forced displacement girl for the two cases means for the force applied by student A and by student B is giving here in the problem like this. First of all, four student A. The force is constant and that force and the forces being represented over the Y axis, which is measured in newton. And the displacement means a position is being represented on the X axis and that is measured in nature. This constant force here is 30 newton. Now, the force applied by the student B was initially having um agreed Europe 14 Newton, after which it starts decreasing. And it is given like this here in the problem. So this is 14 U. Turn over here. Then if you look for the the stands here, at which these two curves intersect each other, this position is given us to meter. So we conclude that for a decline of force of 10 newton, This decline in the force applied by student B takes place or a displacement of two m. So, using unitary methods, suppose the work done by the two students becomes equal after a displacement of X. So up to this displacement of X. What will be the drop in the force applied by student be. We have to find this drop. So to find this, we will use unitary matters which says four, our distance of two m. The reduction in force is dan newton. So For a distance of one m, this reduction in the force applied by Student B will be 10 mountain by two m is this is five newton. So what a distance of X meter. This reduction will be five into X. Newton means here, the force will be reduced by five X. So remaining force after distance X. The remaining force applied by Student B will be 40 minus five X. So At this distance, if we look for the two areas, one Is that covered by Student eight. Here, this is this is a rectangle so we can see over our distance. X. Work done by the student mm is W. E. And that is given simply by the area of rectangle shown in green and that is X means land. And here this width, Which is 30. So we can see this is Turkey ex jews. Similarly work done by student B is W. B. And that will be given by the area so far, trapeze iem here, this is doctorate museum. This And the area of Trumpism is given by the summer of parallel sides. Here, the parallel side, this parallel site is measured to be 14 Newton And this parallel side will be measured to be 40 -5 x. And the distance between them is X. So here it becomes half into some of parallel sites, 40 plus 40 minus five X. Multiplied by distance between them which is X. So it comes out to be AT -5 X. Into X x two Jews. But as these two works are the same. So these 30 x. is equal to 80 -5 x Into X, divided by two, cancelling this X. And making a cross multiplication, we get 60 is equal to 80 minus five X. Or we can say five X equals to 80 minus 60 since this is 20. So This X comes out to be four metre means after moving four metre, the work done by both, the students will be equal. Hence, here, we can say our option. See is correct. Here. Thank you.

And this problem, you are given a force distance graph, and it tells you that one student applies a constant force of 30. Newton's the entire time, and the other student starts off pushing at 40 Newton's but then grows tired and eventually reduces its force overtime so we wouldn't calculate or determine when the work done by each of the students is the same. So from a position are forced position graph. The area under a forest distance graph is equal to the work done. So if I look at my graph, I can visually see that this little segment fits in to this little segment. So if our to imagine this as having kind of flipped and scooted into that little space, the student who was pushing at 40 Newton's has done this amount off work. They can actually calculate that value, but it actually will be necessary. So the student who was pushing at 40 has done this green amount of work where, as a student who was pushing 30 because he was pushing that 30 the entire time, he's also done this same amount of work, so at position for the area's under the graph are the same for both students. So the work done is the same for both students at position for

Who in problem number 39 we have a athlete doing the push up. Our fleet is two meters tall and his weight. Where's masses? 82 kilograms and it's located center Mass is located 1.15 from his feet. So at 82 kilograms, that makes his weight. 803.6 Newtons. Now he has two hands. Here are two feet. Sorry. So we're gonna call this his force of his feet and he has two hands located here. They would like to know the force of his feet on. Find the force on each foot and on each Hagen. That's assuming that they, each hand and foot are exert equal forces. So we're gonna look at this two ways. First, we're gonna look at the some of the torques we know those air zero. Yeah. We're gonna look at the sum of our forces in the UAE direction and we know those air zero. No, looking at our talks to begin with, we know for pivoting a defeat. We know that that lever arm zero. So we have two forces that provide torque. It's the weight in the hands. Okay, So our weight 803 0.6. It's lever arm. 1.15 would equal the Twerk, you know, doing both hands that they're too f ages. And that is located 1.7 from the pivot on solving this Ferrari. I'm sorry for F h. We get each hand is supporting 272 mutants. We're here looking at our forces in the direction. We would know that two feet and two, it was two hands. We're gonna have to hold up his weight of 803.6. We do now. Already. Our hands force of 2 72 substituting in in solving four feet. What would have to support 100 and 30? It's Thank you for


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