Question

Please Help, Easy medal:

What is the maximum work you will do to move the box from A to B?

I am having trouble finding the hypotenuse length, then I can do the rest by myself

Answer 1

divide the paths into pathlets and find work 4 each and then find their sum

Answer 2

no no no, I understand how to find them, I have already done that, but I am trying to create the BEST path which I created a triangle, and in order to find the distance of that path I have to find the hypotenuse of that triangle that I made,, but I cant remember how to get it.

Answer 3

Pythagoras theorem L3 is hypotenuse .[L3]2=[L2]2+[L1]2

Answer 4

i dont understand what u typed

Answer 5

oh okay, nvm

Answer 6

Restating haseebgul17,\[L_3^2=L_1^2+L_2^2\\\qquad\Downarrow\\L_3=\sqrt{L_1^2+L_2^2}\]

And that path would yield the least amount of work you could do to get from A to B. The maximum work you can do is infinite, or however far you can push the box before you.. err.. stop...

Answer 7

That's assuming that there is friction.

Otherwise, work will be the same no matter how you get there.

Answer 8

awesome! thank you

Answer 9

You're welcome! Any questions? Anything that didn't make sense?

Answer 10

so let me see if I got it right, \[L _{3= } \sqrt{2^{2} + 3^{2}} \]

Answer 11

\[L _{3}= 3.6\]

Answer 12

so the work done is: \[-(0.26)(3.7kg)(9.81 m/s ^{2})(6.6m) \]

Answer 13

= -33.97

Answer 14

is that correct?

Answer 15

Let me think.. So the box is being lifted? No friction?

Answer 16

Wait... I'm guessing \(.26=\mu_k\), right?

Answer 17

the problem says: "calculate the work done by friction as a 3.7kg box is slid along the floor from point A to point B in the figure below along paths 1, 2, 3. assume the coefficient of kinetic friction between the box and the floor is .26. AND i solved all the work done for each path, and my professor assigned for bonus points "What is the minimum work you will do to move the box from A to B... and that is what i am solving now. understand?

Answer 18

Ah! I understand!
Well, you said \(L _3 =\sqrt{2^{2} + 3^{2}}\), but shouldn't that be \(L_3 =\sqrt{3^{2} + 3^{2}}\approx 4.2\)? You wrote that in your picture.

Answer 19

Nvm, you're right!

Answer 20

i am right?

Answer 21

About the \(L_3\)! I'm about to plug the rest into the calculator.

Answer 22

Yeah! You made a typo in the calculation, but you got the right answer! You type 6.6m instead of 3.6m for \(L_3\).

Answer 23

typed*

Answer 24

Now, you do positive work, because your force is in the positive direction. Just because you push it forward.

Answer 25

oh my bad, sorry about my mistype,

Answer 26

The friction force is opposite in direction to the distance, so its work is negative.

Answer 27

Haha, np. I mistype too much myself.

Answer 28

but I thought \[Work _{by you } = -W _{friction}\], that is waht your professor said

Answer 29

Yup!
It's tricky.
So, you found the work done by the friction force, but you had a positive 3.6m. You also had a positive friction force, which was \(9.81\ [m/s^2]\ 3.7\ [kg]\).

Answer 30

right......

Answer 31

Well, lets look at this simple diagram, where a box is moving to the right on a floor that has friction.

Answer 32

The distance you move is opposite the friction force applied.

So make the direction of movement be positive 3.6m, while the friction force is negative (9.81)(3.7) [N].