Calculating Force Exerted on a Block in a Vertical Lift

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The discussion revolves around calculating the force exerted on a 40.0 kg block being lifted by a winch in a wooden hanger. The winch performs 2.08 x 10^4 J of work, and participants clarify that the relevant equation is work done equals force times distance. It is noted that the height of the lift is not 300 m but rather the maximum height of 52.0 m. Participants emphasize the importance of understanding potential energy and gravitational force in this context. The conversation highlights the need for clarity on the formulas and values to accurately determine the force exerted on the block.
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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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