# Find Speed of Blocks w/ Pulley & Spring

• BSCS
In summary, when the 30 kg block is 20 cm above the floor, the 20 kg block has a minus sign in its Ug point.
BSCS
A 20.0 kg block is connected to a 30.0 kg
block by a string that passes over a light frictionless
pulley. The 30.0 kg block is connected to a spring that
has negligible mass and a force constant of 250 N/m,
as shown in the figure. The spring is unstretched
when the system is as shown in the figure, and the
incline is frictionless. The 20.0 kg block is pulled 20.0
cm down the incline (so that the 30.0 kg block is 40.0
cm above the floor) and released from rest. Find the
speed of each block when the 30.0 kg block is 20.0 cm
above the floor (that is, when the spring is
unstretched).

The solution says to take the lowest point reached on the ramp by the 20.0 kg block as Ug = 0.

Then, it takes Ug of the 30.0 kg block to be based on +20.0 cm... I don't understand this. Isn't the y-distance from the lowest point on the ramp of the 20.0 kg block to the highest point of the 30.0 kg block *not* 20.0 cm?

That sounds very strange, could you show the solution or an answer and what your problem in solving the question is?

$$\Delta U_{g} = (m_{2}sin\theta-m_{1})gx$$

$$\Delta U_{g} = [20.0 kg sin40 - 30.0 kg](9.80)(0.200) = -33.6 N$$

BSCS said:
Isn't the y-distance from the lowest point on the ramp of the 20.0 kg block to the highest point of the 30.0 kg block *not* 20.0 cm?

Why would this distance be of any interest? For Ug, all you would be interested in is the change in the height of each block.

So, for the block on the incline, that would be the change in vertical distance as it travels back up the incline, and for the hanging block, it would just be 20 cm.

Or am I nuts?

Dorothy

Dorothy Weglend said:
Why would this distance be of any interest? For Ug, all you would be interested in is the change in the height of each block.

So, for the block on the incline, that would be the change in vertical distance as it travels back up the incline, and for the hanging block, it would just be 20 cm.

Or am I nuts?

Dorothy

I can't comment on your sanity , but that goes back to my problem... How can it be 20 cm for the hanging block if Ug is not 0 at the point we would expect?

Well, the hanging block would lose graviational energy, as the block on the incline would gain gravatational energy...

You have the solution, I guess? I get 1.24 m/s... Is that right?

(Just trying to confirm my sanity, I guess)

Dorothy

BSCS said:
I can't comment on your sanity , but that goes back to my problem... How can it be 20 cm for the hanging block if Ug is not 0 at the point we would expect?
The blocks have different Ug points. As what matters, as Dorothy said, is the difference in height between the two states. Having one common Ug point would make matters more complicated. So the Ug point for the first block is yes at the lowest point but for the other block it is at the top of the spring thus it will become 20cm.

Dorothy Weglend said:
Well, the hanging block would lose graviational energy, as the block on the incline would gain gravatational energy...

You have the solution, I guess? I get 1.24 m/s... Is that right?

(Just trying to confirm my sanity, I guess)

Dorothy

1.24 m/s is correct.

ponjavic said:
The blocks have different Ug points. As what matters, as Dorothy said, is the difference in height between the two states. Having one common Ug point would make matters more complicated. So the Ug point for the first block is yes at the lowest point but for the other block it is at the top of the spring thus it will become 20cm.

Ok, this makes sense. So, I can have a separate Ug = 0 point for each element in the system?

BSCS said:
Ok, this makes sense. So, I can have a separate Ug = 0 point for each element in the system?
Indeed, it is just a matter of definition.

As all you are using is the difference in height, thus your definition of the ug point does not matter. Say your Ug point was on Mars and then the block was raised by 20cm.

difference in potential energy: mgh2-mgh1=mg(h2-h1)=mg(dmars+20-dmars)=mg*20

For me, the key to understanding these things it to think of the system energy. In the present case, the 20 kg block will have zero gravitational energy, but the 30 kg block will not, it will have some Ug.

When the system moves, then the 20kg block will gain Ug, as the 30 kg block will lose it. In other words, for the final state of the system, you will have a minus sign in front of the term with the 30 kg in it.

## 1. How does a pulley affect the speed of blocks?

A pulley can change the direction of the force required to move the blocks, but it does not change the speed at which the blocks move. The speed of the blocks is determined by the force applied and the mass of the blocks.

## 2. How can the spring affect the speed of the blocks?

The spring can act as a force on the blocks, either pushing or pulling them, which can increase or decrease their speed depending on the direction of the force and the spring constant.

## 3. How do you find the speed of blocks with a pulley and spring?

To find the speed of the blocks, you will need to measure the force applied to the blocks, the mass of the blocks, and the spring constant. Then, you can use the equation v = √(2F/m + k/m) to calculate the speed of the blocks.

## 4. Can the speed of blocks with a pulley and spring change over time?

Yes, the speed of the blocks can change over time if the force applied or the spring constant changes. Additionally, friction and other external forces can also affect the speed of the blocks.

## 5. How does the angle of the pulley affect the speed of the blocks?

The angle of the pulley does not directly affect the speed of the blocks. However, it can change the direction of the force, which can affect the speed of the blocks if the force is not applied in the same direction as the blocks are moving.

• Introductory Physics Homework Help
Replies
33
Views
3K
• Introductory Physics Homework Help
Replies
14
Views
2K
• Introductory Physics Homework Help
Replies
52
Views
1K
• Introductory Physics Homework Help
Replies
27
Views
6K
• Introductory Physics Homework Help
Replies
1
Views
3K
• Introductory Physics Homework Help
Replies
10
Views
305
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
18
Views
4K
• Introductory Physics Homework Help
Replies
1
Views
4K