How Fast Will Box A Hit the Floor in a Pulley System?

  • Thread starter Thread starter Johnny_99
  • Start date Start date
  • Tags Tags
    Energy
AI Thread Summary
The discussion revolves around calculating the speed of Box A when it hits the floor in a pulley system involving two boxes with different masses. The initial potential energy of Box A is calculated, but the mistake occurs by not accounting for the kinetic energy of Box B, which is also in motion. The conservation of energy principle is emphasized, indicating that the total energy change must consider both boxes. The relationship between the speeds of the two boxes, connected by a rope, is crucial for solving the problem correctly. Understanding these energy dynamics is essential for accurately determining the final speed of Box A upon impact.
Johnny_99
Messages
3
Reaction score
0

Homework Statement


Two boxes are connected over a pulley and hel AT REST. Box A has a mass of 15kg and box B has a mass of 12kg. If the bottom of box A is originally 85cm above the floor, with what speed will it contact the floor when the system is released? Use conservation of energy and assume that friction is negligible. (Hint: Think abou the total energy of the system before and after the block A is released)


Homework Equations


ET= mgh + 1/2mv squared
Ek=1/2mv squared
Ep=mgh


The Attempt at a Solution


Before released: Eta= (15)(9.8)(.85)
Eta= 124.95J
After released: Etb= mgh
Etb= 12(9.8)(.85)

up to this point, the test says I have it write (as seen from the teachers marking)

then this is where i somehow went wrong: 124.95-99.96= Change in Energy
*at this point, the marker said, "B HAS Ek TOO!"

then i did: 24.99= Change in Energy
24.99= 1/2mv squared
24.99= 1/2(15)v squared
3.332m/s=v
3.3m/s=v

This is unfortunately not the correct answer. If anyone can assist me in where i went wrong that would be great!
 
Physics news on Phys.org
I think block B is still moving when A hits the floor. That's probably what your teacher was talking about (there's no floor to stop B).

So

U_A = K_A+K_B+U_B

Dorothy
 
what is that?

what does U stand for Dorothy?
 
Not only is B moving, but what is the relationship between the speeds of A and B? (They are connected by a rope!)

A useful way to view conservation of energy is in terms of changes:
\Delta{E} = \Delta{KE} + \Delta{PE} = 0

Remember: You want the change in energy of the system--both boxes.

(Dorothy is using U to stand for potential energy.)
 
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