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Problem with acceleration

  • Thread starter Ugnius
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Moved from technical forums, so no template
Hi , im trying to practice working with new exercises and i've met some problems.
Lifting object with mass of 2kg to height of 10m we do 240J mechanical work.
What is acceleration of lifting?

My calculations:
A=Fs=Fh
F=A/h = 240/10 = 24N

a= F/m = 24/2 = 12 m/s^2 , somehow it's not the right answer
 
70
3
There is a downward mg which compensates the mechanical work done?

your answer is correct if there is no mg force

in fact,part of the mechanical work done is to compensate the work done by the mg force
 
24
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No , that's all what's written in question
 
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The word "lifting" implies gravity.
 
70
3
Lifting object with mass of 2kg to height of 10m we do 240J mechanical work.
What is acceleration of lifting?

the final answer is 2 ?
 
24
0
2 m/s^2 yes
 
70
3
in fact

your answer is correct if there is no gravity.
 
70
3
if there is gravity
then you will feel a "resistance" from the gravity

part of your 240J will try to compensate this "resistance" so that it is impossible that the force that you provide will be the same of the case of no gravity force
 
24
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Lifting object with mass of 2kg to height of 10m we do 240J mechanical work.
What is acceleration of lifting?

the final answer is 2 ?
How did you get 2? can you write the equation?
Was it a=F/m - g?
 

haruspex

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I'm unsure how to interpret the question.

If this is in a terrestrial setting, we can calculate that part of the work that went into overcoming gravity, i.e. into achieving the PE gain. That is less than 240J. Where did the rest go? Since it asks about acceleration, maybe we are supposed to assume there is residual KE, i.e. it is still moving upwards. If so, we can compute the gained velocity, but without knowing how long this took there is no way to find the acceleration.

Alternatively, this is extraterrestrial, and all they want you to find is the local gravitational acceleration, but then that would be 12m/s2, not 2.
 

NascentOxygen

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If so, we can compute the gained velocity, but without knowing how long this took there is no way to find the acceleration.
The object was lifted 10m.
 

haruspex

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Yes, although it isn't stated, I think maybe the solution should assume constant acceleration.
That does seem to be the most likely error/omission in the question.
 

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