Confused conceptually about work done with pulleys

[BOAS]

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 20.0N block is the weight of the 12.0N block. The work done on the 20.0N block is therefore F x d = 12 x 0.75 = 9 Nm

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

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

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

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

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 f_{k} = \mu_{k} N = 6.5N, so the net force when the box is moving is 5.5N, and thus the work done on the 20N box is w = f x d 4.125N.

I think the work done on the 12N box is still -9Nm 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!

 

[Vibhor]

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

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

I think this is correct, to expect the net work done to be 0, 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.

 

[BOAS]

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 0.76 N, by using Newton's second law and simultaneous equations. The work done on the 20N block is therefore w = f*d = 0.57Nm

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

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

Thanks for your help.

 

[Vibhor]

Thanks for the response.

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

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

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 .

 

[BOAS]

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.

 

[Vibhor]

Using 20/g and 12/g as masses should give you T = 7.5N .

 

[BOAS]

∑F_x = 20/g * a = T

∑F_y = 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

 

[Vibhor]

∑F_x = 20/g * a = T

∑F_y = 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 .

 

[BOAS]

Ah - you are correct, though I'm getting 3.68ms^-2, using g=9.81ms^-2.

Algebra error. Thanks!