Calculating total work using kinetic-work theorem

  • Thread starter Thread starter tjw137
  • Start date Start date
  • Tags Tags
    Theorem Work
AI Thread Summary
The discussion revolves around calculating the total work done by a boy and a girl on a frictionless pond after they push off each other. The girl, with mass 2m, slides to the left at speed v, while the boy, with mass m, moves to the right. The total kinetic energy (KE) is initially miscalculated as 3/2(mv^2), but it should be 3mv^2 due to the girl's mass being 2m. The correct approach involves using conservation of momentum to determine the boy's speed based on the girl's known speed. Clarifications were made regarding the mass and kinetic energy calculations, leading to a better understanding of the problem.
tjw137
Messages
4
Reaction score
0
1. A boy of mass m and a girl of mass 2m are initially at rest at the center of a frozen pond. They push each other so that she slides to the left at speed v across the frictionless ice surface and he slides to the right as shown above. What is the total work done by the children.



2. KE=1/2(mv^2)



3. I was thinking of adding the kinetic energies together. For the girl, I was thinking mv^2 and 1/2 mv^2 for the boy. This would give me 3/2(mv^2), but the correct answer should be 3mv^2. I need help understanding this concept, and how you would arrive to this conclusion. Thanks!
 
Physics news on Phys.org
You are given that the speed of the girl is v, so how her KE be mv^2?

But first things first. How can you figure out the speed of the boy?
 
By conservation of momentum! Thanks for pointing me in the right direction.
 
Doc Al said:
You are given that the speed of the girl is v, so how her KE be mv^2?
But you were correct about that! (Her mass is 2m, not m. :redface:)

tjw137 said:
By conservation of momentum!
Exactly!
 
I had written her mass in terms of the boys mass. I was not clear on that in my initial post. Sorry about that. I should have used subscripts.
 
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