What is the Work Involved in Lifting and Moving a Box with a Crane?

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The discussion focuses on calculating the work involved in lifting and moving a box using a crane, with specific variables provided: weight (G=8400N), height (h=35m), and horizontal distance (d=10m). The work is divided into two components: W1 for lifting and W2 for horizontal movement, with the formulas W1=F1*h and W2=F2*d. The lifting force (F1) is determined by the weight of the box, while the horizontal force (F2) is equal to the mass times acceleration, but the acceleration is not specified. The conversation emphasizes the need to assume slow lifting to simplify calculations, indicating that the lifting force is slightly greater than the weight to achieve elevation. Understanding these principles is crucial for solving the exercise.
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1. The problem statement, all variables and given/known
G=8400N
h=35m
d=10m
g=10m/s^2

A crane elevate a box with G=8400N at h=35. After this, the box is moved horizontal on distance 10m. No friction.

Work = ?
ps. Sorry for bad English.

Homework Equations


I think that W=W1 ( work on elevating ) + W2 ( work on moving ).
W1=F1*h and W2=F2*d

In the first I must tu find F1.
F=F1-G ==> F1=G-F=G-ma

Now, F2:
F=F2==> F2=m*a.

W=F1+F2=G-ma+m*a.

But I don't know a...

The Attempt at a Solution


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I know I am bad but I want to learn. What can I do to solve this exercice ? Thank you very much !
 
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Hello Adrian, :welcome:

Assume the lifting goes very slowly. So the only force that is needed for that is ##mg## = G . Keep the direction of the force in mind.
(why do I say that?).
 
Because F = -G ? Or.. F is a little bit stronger that G to lift ?

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Last edited:
//Sorry for x2 post.
 
Adrian379 said:
Because F = -G ?
That is what is meant, yes.
 
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