# Confused conceptually about work done with pulleys

1. Oct 28, 2014

### 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. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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!

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2. Oct 28, 2014

### Vibhor

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. Oct 28, 2014

### BOAS

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.

Last edited: Oct 28, 2014
4. Oct 28, 2014

### Vibhor

This is incorrect .Recheck your calculations.

I see you have edited your response . Your initial response in post#3 was correct .

Last edited: Oct 28, 2014
5. Oct 28, 2014

### BOAS

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. Oct 28, 2014

### Vibhor

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

7. Oct 28, 2014

### BOAS

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. Oct 28, 2014

### Vibhor

I think it should be a = 3.75 ms-2 .

9. Oct 28, 2014

### BOAS

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

Algebra error. Thanks!