Help solving for energy please?

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The discussion centers on calculating the work done during a squat involving a person weighing 50 kg who lifts an additional 100 kg. Initially, the participant calculates the force using F=ma, resulting in 500 N, and computes work as W=F×d, yielding 600 J. Confusion arises regarding whether to include the additional weight of 100 kg, with suggestions to add it to the lifter's mass, leading to a total of 150 kg. The participant mistakenly multiplies the work by 2, believing it accounts for both the downward and upward motion of the squat, but this assumption is questioned. Ultimately, the correct approach requires considering the total weight lifted without unnecessary multiplications, emphasizing the importance of accurately applying physics principles.
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A screenshot of the question can be found at the bottom of this post.1. Homework Statement

D's mass is 50kgs
He does full squats over 100 kgs
His displacement is 1.2 m
acceleration=10

Homework Equations


F=ma
W=(force)(displacement)

The Attempt at a Solution


First I need to transform his weight from kg's to Newtons.
His weight in Newtons is 490.3[/B]
Now I can solve for work
W=(490.3)(1.2)
W=588.36

This is not a possible answer choice on the sheet,may I please have some help understanding what I have done wrong?
 

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You've accounted for his weight, but what about the 100 kg?

And they suggest g=10 m/sec/sec, so you don't need to be so precise and you could do it with mental arith.
 
Merlin3189 said:
You've accounted for his weight, but what about the 100 kg?

And they suggest g=10 m/sec/sec, so you don't need to be so precise and you could do it with mental arith.

I wasn't sure if I should convert his squatting weight to Newtons,but in that case it would be 980.6 Newtons
 
To me the language is strange, because I thought squatting was intransitive. But the question only seems to make sense if they mean he is squatting while holding 100kg.
So if he is 50kg and is holding 100kg, when he squats and stands again, what is the total weight he lifts?

Your conversions are correct, though as I said, they don't ask you to use g= 9.806, so you could just use g=10 msec-2
 
Merlin3189 said:
To me the language is strange, because I thought squatting was intransitive. But the question only seems to make sense if they mean he is squatting while holding 100kg.
So if he is 50kg and is holding 100kg, when he squats and stands again, what is the total weight he lifts?

Your conversions are correct, though as I said, they don't ask you to use g= 9.806, so you could just use g=10 msec-2

So would I add the two masses together 50kg + 100kg=150 kg which translates to 330.7 lbs. From there should I multiple by the gravity and then divide by the displacement?
 
Physicsis said:
I add the two masses together 50kg + 100kg=150 kg
Yes.
Physicsis said:
translates to 330.7 lbs
Why ever would you want to do that? What are the units in the answer options?
Physicsis said:
From there should I multiple by the gravity
Yes.
Physicsis said:
then divide by the displacement?
What standard equation or physical law leads to that?
 
haruspex said:
Yes.

Why ever would you want to do that? What are the units in the answer options?

Yes.

What standard equation or physical law leads to that?

Sorry on the dividing by the displacement theory I was just being stupid,forgive me. I think I've figured out what to do though. Notice on the picture I posted of the equation it is mentioned that F=ma therefore I multiplied his mass by 10. F=500N and then from there I multiplied by 1.2 and got an answer of 600 then I had to multiple 600 by 2 and in result I got 1200. In this question I am trying to find out how much energy the gentlemen burned doing a full squat.
 
Physicsis said:
Sorry on the dividing by the displacement theory I was just being stupid,forgive me. I think I've figured out what to do though. Notice on the picture I posted of the equation it is mentioned that F=ma therefore I multiplied his mass by 10. F=500N and then from there I multiplied by 1.2 and got an answer of 600 then I had to multiple 600 by 2 and in result I got 1200. In this question I am trying to find out how much energy the gentlemen burned doing a full squat.
So what is your final answer (include units)?
 
haruspex said:
So what is your final answer (include units)?
My final answer is 1200 (J), can I have your help with a different question if you don't mind?
 
  • #10
Physicsis said:
My final answer is 1200 (J), can I have your help with a different question if you don't mind?
That's not right. You've forgotten something. if you cannot see your error, please post all your steps in detail.
Please start a new thread for a new question.
 
  • #11
haruspex said:
That's not right. You've forgotten something. if you cannot see your error, please post all your steps in detail.
Please start a new thread for a new question.
Here are my steps as follows

F=ma
F=(50)(10)
F=500N

W=(force)(displacement)

W= 500*1.2=600

To find joules burned I multiply by 2 I believe

600*2=1200

Final answer 1200(j)
 
  • #12
Physicsis said:
To find joules burned I multiply by 2
Why?
What about the 100kg burden?
 
  • #13
haruspex said:
Why?
What about the 100kg burden?

Well if I were to add that into the equation,I would get an answer that is not at all one of the answer choices. The answer choices for this particular problem are

600,1200,1800,and 1000 (all in Joules)
 
  • #14
Physicsis said:
Well if I were to add that into the equation,I would get an answer that is not at all one of the answer choices. The answer choices for this particular problem are

600,1200,1800,and 1000 (all in Joules)
I agree with one of those. Trying to match one of the answers does not justify leaving out the 100kg. You should be looking for some other error.

You did not answer my other question. Why do you multiply the work done by 2?
 
  • #15
haruspex said:
I agree with one of those. Trying to match one of the answers does not justify leaving out the 100kg. You should be looking for some other error.

You did not answer my other question. Why do you multiply the work done by 2?

Well when you do a squat you go from the first position and down into the second one, so when I multiplied 500 by 1.2 and got 600 I then multiplied it by 2 and got 1200.
 
  • #16
Physicsis said:
Well when you do a squat you go from the first position and down into the second one, so when I multiplied 500 by 1.2 and got 600 I then multiplied it by 2 and got 1200.
Do you do the same work going down as coming up?
 
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  • #17
haruspex said:
Do you do the same work going down as coming up?
In this case I'd assume so.
 
  • #18
Physicsis said:
In this case I'd assume so.
I can't think why. I would find it much easier to lower 100kg than to raise 100kg.
 
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