What is the speed of Betty on a playground swing?

  • Thread starter Thread starter sonic91
  • Start date Start date
  • Tags Tags
    Energy Motion
AI Thread Summary
Betty, weighing 427 N, swings from a height of 0.88 m to 0.22 m, with gravity at 9.8 m/s². The potential energy at the higher point converts to kinetic energy at the lowest point. Initial calculations yielded 281.82 Joules for potential energy, leading to an incorrect speed of 11.3737 m/s. A correction noted that the weight should be 427 N, not 527 N, and emphasized the need to take the square root in the final step. The problem was resolved with these adjustments.
sonic91
Messages
9
Reaction score
0

Homework Statement


Betty weighs 427 N and she is sitting on
a playground swing seat that hangs 0.22 m
above the ground. Tom pulls the swing back
and releases it when the seat is 0.88 m above
the ground.
The acceleration of gravity is 9.8 m/s2 .
How fast is Betty moving when the swing
passes through its lowest position?
Answer in units of m/s.


Homework Equations


KE=1/2 M V^2
PEg=Mgh
i might be missing an equation here


The Attempt at a Solution


i was thinking that potential energy gravitational would turn into Kinetic energy then i solved for v.

PEg= 427 Newtons*.66 i used the change in height for h and got 281.82 Joules for my answer. then i said that should transfer to KE.

281.82 joules= 1/2* 427/9.8 Kg * V^2 and i got 11.3737 m/s as my answer. when i checked this it turned out to be wrong. i think i may have to apply circular motion concepts to this problem, but I'm not quite sure how to do that.

thanks in advance.
 
Last edited:
Physics news on Phys.org
Your logic looks sound; it must just be a math error throwing you off... If the work you have here is what you typed into your calculator, then I have two suggestions: first, the weight in the 2nd equation should be 427, not 527... secondly, did you take the square root of both sides at the end? Other than that, I don't know what to tell you.
 
yeah I've got it figured out now thanks.
 
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