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BOAS

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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.

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.

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!

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 Statement1. 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!