Calculating Kinetic Energy Using Uniform Velocity and Energy Equations

AI Thread Summary
Avdol is identified as the force doing work when lifting a 35 kg box to a height of 2.0m, while gravity also does work during the process. The discussion highlights confusion regarding the calculation of kinetic energy when the box is dropped, with participants emphasizing the need to understand the relevant physics concepts. A participant calculates the velocity just before impact using the formula √2gh, resulting in a kinetic energy of 690 J, but questions the clarity of the problem's wording. Ultimately, it is suggested that both Avdol and gravity do work, and a more direct approach to calculating kinetic energy is to use the equation Ek = mgh. Understanding these principles is crucial for solving similar physics problems effectively.
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Homework Statement
Avdol lifted a 35 kg box at uniform velocity from the ground up to a height of 2.0m.

a) Jotaro commented Avdol did work. Polnareff says gravity did work. Who is correct? Explain?

b) If Avdol then accidentally dropped it. How much kinetic energy will there be right before it lands? Explain your reasoning. (2 marks)
Relevant Equations
Ek = 1/2 m v^2
For:
a) Avdol did the work because he is the force that is causing the displacement, right?

b) Is there another formula we would have to use? I am confused at how this would work out and what the answer would be.
 
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UnknownQuestioner said:
Homework Statement:: Avdol lifted a 35 kg box at uniform velocity from the ground up to a height of 2.0m.

a) Jotaro commented Avdol did work. Polnareff says gravity did work. Who is correct? Explain?

b) If Avdol then accidentally dropped it. How much kinetic energy will there be right before it lands? Explain your reasoning. (2 marks)
Relevant Equations:: Ek = 1/2 m v^2

For:
a) Avdol did the work because he is the force that is causing the displacement, right?

b) Is there another formula we would have to use? I am confused at how this would work out and what the answer would be.

That's correct for a). For b) where does the kinetic energy come from?
 
The kinetic energy would be the force of gravity pulling?
 
UnknownQuestioner said:
The kinetic energy would be the force of gravity pulling?

Can you think of something other than force?
 
Sorry I am not sure
 
UnknownQuestioner said:
Sorry I am not sure
Regarding this and your other post, the rules are that you must make an effort to solve the problem. We can help but we cannot do the work for you.

You need to have a think about the material you have been studying and what is relevant to a question like this.
 
Oh ok, sorry about that, I will re-attempt and try to do the problem again with more research. Thanks
 
UnknownQuestioner said:
a) Jotaro commented Avdol did work. Polnareff says gravity did work. Who is correct? Explain?

I think I can tell what the wording is implying, but this is an awful question? Both do work, and the total work is zero. Hmm
 
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I used √2gh = v to get velocity (6.3 m/s) then used that in Ek = 1/2 m v^2 to get a final kinetic energy of 690 J. Would this be the correct way to do this?
 
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etotheipi said:
I think I can tell what the wording is implying, but this is an awful question? Both do work, and the total work is zero. Hmm
Agreed, the intended answer is that both are right. But it is unclear whether the total work is zero. We are not told that Avdol stops it at that height, or whether we are to include the work done to accelerate it to the given speed.
 
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  • #11
UnknownQuestioner said:
I used √2gh = v to get velocity (6.3 m/s) then used that in Ek = 1/2 m v^2 to get a final kinetic energy of 690 J. Would this be the correct way to do this?
It is one way, but a bit roundabout. Combining those equations, Ek = 1/2 m v^2 = 1/2 m (√(2gh))^2 = mgh. How could you have got that straightaway?
 
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