Calculating Work in a Pulley System

[james brug]

Work problem--blocks,pulley

Homework Statement

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

Find the total work done on 20.0N block if there is no friction between the table and the 20.0N block.

Homework Equations

$$w=f\cdot s$$
$$f_k=\mu_k\cdot n$$

The Attempt at a Solution

w=12N(.75m)=9J--mastering physics says it's wrong.

 

[w3390]

Ya, your attempt is not correct. What you need to do is find out the force the hanging block exerts on the sliding block. Then you can do W=F*d using this force over a distance of 75cm.

 

[Delphi51]

First, are you sure the blocks are in Newtons? Very odd - would expect kg.
Second, I think the whole thing is more complicated because the blocks are accelerating. A force of 12 N down on the 2nd block will be opposed by the Tension in the string, so
sum of forces = ma
mg - T = ma
T = mg-ma and it will not be the full 12N pulling on the other block.
I think the first step will be to calculate the acceleration of the whole system.

 

[james brug]

Second, I think the whole thing is more complicated because the blocks are accelerating. A force of 12 N down on the 2nd block will be opposed by the Tension in the string, so
sum of forces = ma

That's probably it. There is a problem nearly identical to this in the book, but it says the blocks are moving at a constant speed.

 

[w3390]

I believe Delphi51 is correct.

 

[james brug]

T would give me the force on the 20N block and thus the work, yes?
a=12N/total weight?

 

[w3390]

I think you mean that a=(Force)/(mass) not weight. For the force, use the T from your first statement.

 

[james brug]

What first statement?

 

[w3390]

T would give me the force on the 20N block and thus the work, yes?

This statement. You say that solving for T would give you the force you need in the equation to find work.

 

[james brug]

What about the work done on the other block? Is it the same?

 

[w3390]

The work will not be the same for the other block. In this case the only force acting on it is mg. Therefore, your work will just be mg*.75m and I think it will be negative.

 

[james brug]

I had already found that out, but thanks anyway.