Work in Pulley System: What is the Total Work Done on the 20.0-N Block?

[elsternj]

Homework Statement

Two blocks are connected by a very light string passing over a massless and frictionless pulley (Figure ). The 20.0-N block moves 75.0cm to the right and the 12.0-N block moves 75.0cm downward. Find the total work done on 20.0-N block if there is no friction between the table and the 20.0-N block.

Homework Equations

W=f*s
F=ma

The Attempt at a Solution

Well the only force on the 20N block in the x direction is the tension. The tension is equal to 12N. I would think that the work would be W=12N*.75m. But this is not the case.
A different approach I took was
F_x=ma.
12N = 2.04a
a = 5.88
V_x²=2a(x₀-x)
V_x²=2(5.88)(.75)
V_x = 2.97
then used 1/2mv² which also didn't come up with the right answer.
please help this one seems like it should be simple but I am getting lost!

 

[elsternj]

scratch this. figured it out!

 

[elsternj]

however this part is confusing me:

Find the total work done on 20.0-N block if coefficient of static friction =0.500 and coefficient of kinetic friction =0.325 between the table and the 20.0-N block.

I've never seen a problem where it gives 2 friction coefficients so i am unsure how to set that up. Do I disregard the static friction? i'd imagine that it is there for a reason. I tried finding the new acceleration given the friction using just the kinetic coefficient and then finding the tension and ultimately the work. but the answer wasn't right. any insight on this part of the problem?

 

[SammyS]

What's the exact wording of the problem?

Do the blocks start from rest? If so, you need to determine whether the tension is great enough to overcome static friction. (In this case the tension IS 12 N. --- Why?)

On the other hand, if the wording is similar to that in the Original Post, then use kinetic friction.

 

[elsternj]

The exact wording of the problem is:

Two blocks are connected by a very light string passing over a massless and frictionless pulley (Figure ). The 20.0-N block moves 75.0cm to the right and the 12.0-N block moves 75.0cm downward. Find the total work done on 20.0-N block if coefficient of static friction =0.500 and coefficient of kinetic friction =0.325 between the table and the 20.0-N block.

In the case of static friction the tension would be 12N because there would be no movement, no acceleration. So if it doesn't move then wouldn't work be 0 in that case? I need to find the total work done given these 2 coefficients of friction. i tried to find it using just the kinetic and the answer was incorrect.

 

[SammyS]

OK.

What did you do (specifically) to find the work done when you used the kinetic coefficient?

 

[elsternj]

Okay this is what I did:

Sum of the Forces in the Y direction for block B = mg - T = ma
Sum of the Forces in the X direction for block A = T - f = ma

f_k = 20N*.325 = 6.5

So we need to find T and a.

Sum of the Forces in the Y direction for block B = mg - T = ma

mg_B - ma_B = T .. So

Sum of the Forces in the X direction for block A = mg_B - ma_B - f = ma_A

So my only unknown here is a so I solve this for a
a=1.69m/s

I take this acceleration and plug it in my equation to find T.

mg-ma = T
12 - (1.22)(1.69) = T
T = 9.94

Sum of the Forces in the X direction for block A = T - f = ma

9.94 - 6.5 = 3.44

W = f*s
W = 3.44 * .75 = 2.58

Ah... and what do you know. that's the right answer. Why did they even include static friction coefficient?!

 

[SammyS]

...

Ah... and what do you know. that's the right answer. Why did they even include static friction coefficient?!

I don't know either. Maybe they had a part (c) in mind.