Calculating Force Exerted on a Block in a Vertical Lift

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Homework Help Overview

The problem involves calculating the force exerted on a block being lifted by a winch in a vertical lift scenario. The block has a mass of 40.0 kg, and the winch performs 2.08 x 10^4 J of work to lift it. The context includes concepts of work, force, and gravitational potential energy.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between work done and potential energy, with some questioning the relevance of kinetic energy in this context. There are attempts to clarify the necessary equations and the values needed for calculations.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the use of potential energy, but there is no consensus on the correct approach or the distance involved in the calculation.

Contextual Notes

Participants note that the class has not yet covered potential energy, leading to uncertainty about the relevant formulas. There is also confusion regarding the correct distance the block is lifted, as assumptions about the height of the hanger are questioned.

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Homework Statement



During World War II, 16 huge wooden hangers were built for the U.S. Navy airships. The hangers were over 300 m long and had a maximum height of 52.0 m Imagine a 40.0 kg block being lifted by a winch from the ground to the top of the hanger's ceiling. If the winch does 2.08 x 10^4 J of work in lifting the block, what force is exerted on the block?

Homework Equations



KE = 1/2 mv^2
Net Work = deltaKE

The Attempt at a Solution



Could someone please explain to me, in detail, what to do here?
if you can, much appreciated!
Thank You
 
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welcome to PF!

well, hello duckywucky! welcome to pf! :wink:

(have a delta: ∆ and try using the X2 tag just above the Reply box :smile:)
duckywucky said:
Imagine a 40.0 kg block being lifted by a winch from the ground to the top of the hanger's ceiling. If the winch does 2.08 x 10^4 J of work in lifting the block, what force is exerted on the block?

Net Work = deltaKE

No, ∆KE = 0 (it's not launching the thing, is it? :rolleyes: :smile:)

Try ∆PE. :smile:
 
So does that mean I would do (40.0)(9.81)(2.08 x10^4)?
I came up with 814.192 but in scientific notation, it would be: 8.16 x 10^6 J?
 
duckywucky said:
So does that mean I would do (40.0)(9.81)(2.08 x10^4)?

why?? :confused: explain in words.
 
I have no idea to be honest. Our class hasn't covered potential energy yet so i don't know what the formula is for it. XD
 
duckywucky said:
I have no idea to be honest.

hmm … thought so! :rolleyes:

ok, so presumably the only equation you know is work done = force times distance.

in that case, what figures have you been given that you can put into that equation? :smile:
 
Work done is 2.08 x 10^4.
You're looking for Force, but I'm not sure the distance...maybe 300 m?
 
duckywucky said:
Work done is 2.08 x 10^4.
You're looking for Force, but I'm not sure the distance...maybe 300 m?

Read the question

"lifted by a winch from the ground to the top of the hanger's ceiling" …

that's not 300 m.
 
Work done is 2.08.
Force of gravity is 9.81
And the mass of the block is 40.0 kg.
 

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