Confused conceptually about work done with pulleys

In summary, the conversation discusses a problem involving work done and pulleys. The participants are trying to solve for the work done on the 20.0 N and 12.0 N blocks under different conditions, such as with and without friction. There is some initial confusion and incorrect calculations, but eventually the correct solutions are found using Newton's second law and simultaneous equations.
  • #1
BOAS
552
19
Hello,

I am confused about a problem regarding work done and pulleys. I am confused mainly by the last two, but not 100% confident in the first two either.

1. Homework Statement


Two blocks are connected by a very light string passing over a massless and frictionless
pulley (Figure 2). The 20.0 N block moves 75.0 cm to the right and the 12.0 N block moves
75.0 cm downward.

(image attached)

Find the work done on
(a) the 20.0 N block if there is no friction between the table and the 20.0 N block.
(b) the 12.0 N block if there is no friction between the table and the 20.0 N block.
(c) the 20.0 N block if µs=0.500 and µk=0.325 between the table and the 20.0 N block.
(d) the 12.0 N block if µs=0.500 and µk=0.325 between the table and the 20.0 N block.

Homework Equations

The Attempt at a Solution



For case (a) the force that is moving the [itex]20.0N[/itex] block is the weight of the [itex]12.0N[/itex] block. The work done on the [itex]20.0N[/itex] block is therefore [itex]F x d = 12 x 0.75 = 9 Nm[/itex]

For case (b) I think I can find the work done by finding the change in gravitational potential energy of the [itex]12.0N[/itex] block which is [itex]mgΔh = -9 Nm[/itex].

I think this is correct, to expect the net work done to be [itex]0[/itex], since there is no friction which is non-conservative.

However, for question (c), I am confused about using the coefficient of static friction.

The [itex]20N[/itex] box will move if the static friction force is less than [itex]12N[/itex]. [itex]μn= 10N[/itex] so the net force in the [itex]x[/itex] direction is [itex]2N[/itex].

Now I know the box will move, but is this information relevant to the rest of the question?

The kinetic friction force comes to be [itex] f_{k} = \mu_{k} N = 6.5N[/itex], so the net force when the box is moving is [itex]5.5N[/itex], and thus the work done on the [itex]20N[/itex] box is [itex] w = f x d 4.125N[/itex].

I think the work done on the [itex]12N[/itex] box is still [itex]-9Nm[/itex] because it's change in GPE is still the same, it just changes more slowly.

I know this is a bit of a ramble, but i'd really appreciate it if you could confirm whether my thoughts make sense.

Thanks!
 

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  • #2
BOAS said:
For case (a) the force that is moving the [itex]20.0N[/itex] block is the weight of the [itex]12.0N[/itex] block. The work done on the [itex]20.0N[/itex] block is therefore [itex]F x d = 12 x 0.75 = 9 Nm[/itex]

For case (b) I think I can find the work done by finding the change in gravitational potential energy of the [itex]12.0N[/itex] block which is [itex]mgΔh = -9 Nm[/itex].

I think this is correct, to expect the net work done to be [itex]0[/itex], since there is no friction which is non-conservative.

Both the parts are wrong .I haven't checked other parts yet .

In a) force acting on 20 N block is the tension in the string attached to it ,not the weight of the hanging block .

In b) the change in GPE is not equal to the work done .

Draw FBD of the two blocks and write Newton's 2nd Law separately.
 
  • #3
Vibhor said:
Both the parts are wrong .I haven't checked other parts yet .

In a) force acting on 20 N block is the tension in the string attached to it ,not the weight of the hanging block .

In b) the change in GPE is not equal to the work done .

Draw FBD of the two blocks and write Newton's 2nd Law separately.

Thanks for the response.

I have calculated the tension in the string to be [itex]0.76 N[/itex], by using Newton's second law and simultaneous equations. The work done on the 20N block is therefore [itex]w = f*d = 0.57Nm[/itex]

The net force on the [itex]12N[/itex] block is equal to [itex]12 - T = 11.24N[/itex], thus the work done is [itex]w = f*d = 8.43Nm[/itex]

If this is correct I shall try and apply it to the other parts of the question.

Thanks for your help.
 
Last edited:
  • #4
BOAS said:
Thanks for the response.

I have calculated the tension in the string to be [itex]0.76 N[/itex], by using Newton's second law and simultaneous equations. The work done on the 20N block is therefore [itex]w = f*d = 0.57Nm[/itex]

The net force on the [itex]12N[/itex] block is equal to [itex]12 - T = 11.24N[/itex], thus the work done is [itex]w = f*d = 8.43Nm[/itex]

If this is correct I shall try and apply it to the other parts of the question.

Thanks for your help.

This is incorrect .Recheck your calculations.

I see you have edited your response . Your initial response in post#3 was correct .
 
Last edited:
  • #5
Vibhor said:
I see you have edited your response . Your initial response in post#3 was correct .

Are you sure?

I changed my answer because I used the weight of the block rather than the mass. My new results were found by using 20/g and 12/g in exactly the same equations.
 
  • #6
Using 20/g and 12/g as masses should give you T = 7.5N .
 
  • #7
Fx = 20/g * a = T

Fy = 12 - T = 12/g * a

12 - 12/g * a = 20/g * a

a = 0.375 ms^-2 Sub this value into top equation.

20/g * a = T = 0.76N
 
  • #8
BOAS said:
Fx = 20/g * a = T

Fy = 12 - T = 12/g * a

12 - 12/g * a = 20/g * a

a = 0.375 ms^-2 Sub this value into top equation.

20/g * a = T = 0.76N

I think it should be a = 3.75 ms-2 .
 
  • #9
Ah - you are correct, though I'm getting 3.68ms^-2, using g=9.81ms^-2.

Algebra error. Thanks!
 

Related to Confused conceptually about work done with pulleys

1. What is work done with pulleys?

Work done with pulleys refers to the amount of energy expended in moving an object through a certain distance using a pulley system. It is a measure of the force applied in lifting or lowering an object and the distance it is moved.

2. How is work done calculated with pulleys?

Work done with pulleys can be calculated by multiplying the force applied to the object by the distance it is moved. In a pulley system, the force is equal to the weight of the object being lifted or lowered, and the distance is the height the object is moved.

3. What is the relationship between work and pulleys?

Pulleys are simple machines that make it easier to lift or move heavy objects by reducing the amount of force needed. As the pulley system reduces the force needed, it also increases the distance the object is moved, resulting in the same amount of work being done.

4. Can the work done with pulleys be less than the applied force?

No, the work done with pulleys will always be equal to or greater than the applied force. This is due to the conservation of energy, where energy cannot be created or destroyed, but only transferred or converted from one form to another.

5. How does the number of pulleys affect the work done?

The number of pulleys in a system can affect the work done by changing the direction of the force applied. With more pulleys, the force required to lift an object is divided among them, resulting in a lower force needed to do the same amount of work. However, more pulleys also mean a longer distance is needed to lift the object, increasing the overall work done.

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