• Support PF! Buy your school textbooks, materials and every day products Here!

Two Boxes, a pulley, and KE

  • Thread starter Jared944
  • Start date
  • #1
10
0

Homework Statement



Two boxes are set up in a system as shown in the attached picture.

Box 1 - mass = 5 kg
Box 2 - mass = 2 kg

h = 5m

The system is released, and box 2 hits the ground. With a frictionless system, find the total KE before box 2 hits the ground.

Homework Equations



I am fairly sure I have to use the conservation of energy, which in this case would be:

(1/2)*m*vf^2=m*g*h

((1/2)*m*vf^2)-(m*g*h)=KE


The Attempt at a Solution



I dont quite understand how to go about this problem with the two box system. I can figure out the boxes force and acceleration, but not what I need.

Thanks,
 

Attachments

Answers and Replies

  • #2
radou
Homework Helper
3,115
6
Unfortunately, the attachment doesn't seem to be approved yet. What exactly does 'h' represent?
 
  • #3
10
0
Oh, ok.

Block 1 is sitting on a (frictionless) table attached to block 2 which is hanging off of the table. Both blocks are attached by a rope and a (frictionless) pully.

Before the system is released, block 2 is 5m from the ground (h = 5m)

Thanks for the obervation about the pic...
 
  • #4
radou
Homework Helper
3,115
6
You were right when you said you have to use energy conservation. Since the block are connected with a rope, and since there is no friction, in what relation are the velocities of the blocks?
 
  • #5
10
0
The velocities should be the same. However, Ive calculated the acceleration and the Tension of the blocks in motion. The acceleration and T of block 2 can be illustrated as T-w2=m2(-a)
And Block 1 can be illustrated as 0+T=m1*a

The tension turnes out to be 14 N and the acceleration is 2.8 m/s^2

Im not sure how to use the data...
 
Last edited:
  • #6
radou
Homework Helper
3,115
6
The velocities should be the same. However, Ive calculated the acceleration and the Tension of the blocks in motion. The acceleration and T of block 2 can be illustrated as T-w2=m2(-a)
And Block 1 can be illustrated as 0+T=m1*a

The tension turnes out to be 7.35 N and the acceleration is 2.45 m/s^2

Im not sure how to use the data...
Do you need the acceleration and the tension? Think about how the kinetic energy of the system is defined.
 
  • #7
10
0
So, KE = mgh + w
KE = 2*-9.8*5+14
KE = -84J? How can I reach a negative KE?
 
  • #8
radou
Homework Helper
3,115
6
So, KE = mgh + w
KE = 2*-9.8*5+14
KE = -84J? How can I reach a negative KE?
What exactly is w?

Do you agree if we say that the total kinetic energy of the system equals 1/2 M v^2 + 1/2 m v^2 ?
 
  • #9
10
0
of course.... I was thinking w was the work done by the tension of the system.
 
  • #10
Doc Al
Mentor
44,943
1,206
of course.... I was thinking w was the work done by the tension of the system.
The tension is an internal force and does no net work on the system as a whole. If you were to analyze one mass by itself, then you would need to include the work done by the tension force.

Note also: When calculating changes in gravitational PE using [itex]mg\Delta h[/itex], g is just a positive number (g = 9.8 m/s^2, not -9.8 m/s^2). g is just the magnitude of the acceleration due to gravity.
 

Related Threads on Two Boxes, a pulley, and KE

  • Last Post
Replies
2
Views
942
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
4
Views
3K
Replies
8
Views
4K
Replies
26
Views
1K
Replies
24
Views
5K
  • Last Post
Replies
1
Views
5K
Replies
19
Views
6K
Replies
1
Views
1K
Replies
5
Views
10K
Top