How much total mechanical work does a eight lifter do in 10 repetitions

  • Thread starter Thread starter ShakeWell
  • Start date Start date
  • Tags Tags
    Mechanical Work
AI Thread Summary
The total mechanical work done by an eight lifter performing 10 repetitions of lifting and lowering 200 kg to a height of 1.0m is zero, as the final displacement is zero. Work is defined as the product of force and displacement in the direction of that force, and since the weight returns to its original position, there is no net displacement. While gravity is a conservative force, the muscle force involved is not typically classified as mechanical work in this context. The discussion emphasizes the importance of assumptions made in calculating work and suggests consulting a teacher for clarity. Ultimately, the consensus is that the total mechanical work in this scenario is indeed zero.
ShakeWell
Messages
7
Reaction score
0
ok, tell me if I'm wrong:

the questions asks:
How much total mechanical work does a eight lifter do in 10 repetitions of lifting (and lowering) 200 kg to a height of 1.0m?

I'm thinking, since the final total displacement is zero, the answer is 0 right?
 
Physics news on Phys.org
The "total mechanical work" is the change in potential energy (and here, since the weight is returned to its initial position that is 0) only for a "conservative" force (i.e. no friction). Of course, gravity is conservative but, in general, the force supplied by muscles is not. I don't think it is possible to give a definite answer without knowing precisely what assumptions are made. Exactly what the total mechanical work is depends a lot on how it is moved. Many texts would not call the work the muscles have to do "mechanical" work. While I think there is a very good argument for "0 work", I recommend you discuss this with your teacher!
 


Yes, you are correct. The total mechanical work done by the eight lifter in 10 repetitions would be 0, since the final displacement is zero. This is because work is defined as the product of force and displacement in the direction of the force. Since the weight is being lifted and lowered to the same height, the total displacement is zero and therefore the total work done is also zero.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top