Two objects connencted - finding KE before one reaches the floor

  • Thread starter crazyog
  • Start date
  • #1
crazyog
50
0

Homework Statement


Two boxes are connected to each other as shown. The system is released from rest and the 1.00 kg box falls through a distance of 1.00 m. The surface of the table is frictionless. What is the kinetic energy of box B just before it reaches the floor?

Homework Equations


I believe you should be using an equation such as Ei=Ef
kinetic energy - (1/2)mv^2
and potential energy - mgy


The Attempt at a Solution


so Ei = (1/2)mvi^2 + (1/2)mv^2 + mgy +mgy
= (1/2)(3)(0)^2 + (1/2)(5)(0^2) +(3)(9.8)(0) + (1)(9.8)(1)

Ef = (1/2)mvi^2 + (1/2)mv^2 + mgy +mgy
= (1/2)(3)(vi)^2 + (1/2)(5)(vi^2) +(3)(9.8)(0) + (1)(9.8)(.01)

the velocity would be the same for each since they are connected

or would I do something like
= (1/2)(3+1)v^2 + (1/2)(9.8)(1.0)
solve for velocity...? and then use this velocity to find the KE


thanks for the help!
 

Attachments

  • physics2.doc
    54 KB · Views: 377

Answers and Replies

  • #2
liorda
28
0
next time use imageshack for image posting.

Is there any reference to the height, in the problem? if so - you should consider the potential energy mgh.
 
Last edited:
  • #3
crazyog
50
0
Well there is a hint that says: The gravitational energy will appear as kinetic energy of both boxes. (I'm not sure what this means?) It also says the answer is suppose to be 2.45 J.
I'm not getting anywhere near that.
 
  • #4
konthelion
238
0
1.00 kg box falls through a distance of 1.00 m.
Please label your masses m1 and m2. You are given that the fallen object's mass = 1kg. It does not match with the equations below.
See, you put m=3, m=5, then switch to m=3, m=1. It makes no sense. Label your variables.

so Ei = (1/2)mvi^2 + (1/2)mv^2 + mgy +mgy
= (1/2)(3)(0)^2 + (1/2)(5)(0^2) +(3)(9.8)(0) + (1)(9.8)(1)

Ef = (1/2)mvi^2 + (1/2)mv^2 + mgy +mgy
= (1/2)(3)(vi)^2 + (1/2)(5)(vi^2) +(3)(9.8)(0) + (1)(9.8)(.01)

By the Work-Energy Theorem, [tex]\frac{1}{2}(m_{1}+m_{2})v^2=m_{2}gy[/tex]
where [tex]m_{1}[/tex] is the mass of the box on the table, and the [tex]m_{2}[/tex] is what I'm presuming to be a box hanging from a pulley( I don't see a diagram)
Let y = 1m (the distance it falls)

solve for velocity...? and then use this velocity to find the KE
Yes.
 
  • #5
crazyog
50
0
Yeah I realized that after because I was trying to do the problem while looking at an example that had (3 and 5 kg boxes) sorry!

I solved for velocity and did v= 1.56 m/s

so K= 1/2mv^2 + mgy
K = (1/2)(1)(1.56)^2 + (1)(9.8)(0)

but this does not equal 2.45 J ...what am i missing?
 
  • #6
alphysicist
Homework Helper
2,238
2
Hi crazyog,

I don't think the velocity you solved for is correct; it looks like you might have used m=3kg and 5kg for the kinetic energy terms, instead of m=1kg and 3 kg.

If that's not what happened, please post the numbers you used that gave a speed of 1.56 m/s.

(At this point, your attachment is not approved, so I may be not understanding the problem. However, I did get the answer you said was correct.)
 
  • #7
crazyog
50
0
Yeah I noticed that I wrote the wrong numbers because I was working off of an example and got mixed up, but I don't know how to change that. Sorry!

I used
1/2(m1+m2)v^2=(m2gy)
1/2(1+3)v^2 = (1)(9.8)(1m)
solved for v^2 and got 1.56 m/s
 
  • #8
crazyog
50
0
where do I go from here?
 
  • #9
alphysicist
Homework Helper
2,238
2
Yeah I noticed that I wrote the wrong numbers because I was working off of an example and got mixed up, but I don't know how to change that. Sorry!

I used
1/2(m1+m2)v^2=(m2gy)
1/2(1+3)v^2 = (1)(9.8)(1m)

What I was saying in my last post, is that this equation does not give 1.56 m/s. If you use the wrong masses on the left, (with m=3 and 5), then you get 1.56 m/s, or if you don't multiply by the 1/2 you get 1.56. So I think you either used the wrong masses or had a math error.

Try solving the above equation again, and you'll get a larger answer for the velocity; what do you get?



Also, you can check your answer. For example, if we try plugging in v=1.56 m/s in the above we get for the left side:

(1/2)(1+3) 1.56^2 = 4.8672

which does not equal the right side, so 1.56 is not a solution of the above equation.
 
Last edited:

Suggested for: Two objects connencted - finding KE before one reaches the floor

Replies
10
Views
707
  • Last Post
Replies
7
Views
318
Replies
17
Views
569
  • Last Post
Replies
1
Views
389
Replies
14
Views
488
Replies
11
Views
358
Replies
17
Views
245
Replies
2
Views
310
Top