# Two Boxes, a pulley, and KE

## 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,

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Homework Helper
Unfortunately, the attachment doesn't seem to be approved yet. What exactly does 'h' represent?

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...

Homework Helper
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?

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...

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Homework Helper
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.

So, KE = mgh + w
KE = 2*-9.8*5+14
KE = -84J? How can I reach a negative KE?

Homework Helper
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 ?

of course.... I was thinking w was the work done by the tension of the system.

Doc Al
Mentor
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 $mg\Delta h$, 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.