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Finding massby 2much
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#1
Mar511, 07:01 PM

P: 14

1. The problem statement, all variables and given/known data
You apply a force of .35N up to lift a fork, the resulting acceleration is .15m/s2. What is the mass in grams. Please help I don't know where to start with this simple question. 


#2
Mar511, 07:13 PM

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Start with identifying the net force acting on it, and use Newton's 2nd law. Please show an attempt.



#3
Mar511, 07:18 PM

P: 14

The net force would not be zero and I am only given the applied force. Without a mass I don't know how I can get the force of gravity.
So as far as I can tell I have .35 Fg = m .15m/s2 


#4
Mar511, 07:27 PM

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Finding mass
Let mass = m & use symbols where needed.
What does Newton's 2nd Law say? 


#5
Mar511, 07:46 PM

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0.35 Fg = m .15m/s2
0.35  (m 9.8 m/s2) = m 0.15ms2 The second law says that if the net force is not zero there is an acceleration in the direction of the force. 


#6
Mar511, 07:51 PM

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You apply a force of 0.35 N upward.
If the mass of the for is m, what force is gravity exerting downward on the fork? What is the net force being exerted upon the fork? 


#7
Mar511, 07:56 PM

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Wouldn't I need the mass of the fork (m) to calculate the force exerted by gravity?



#8
Mar511, 07:58 PM

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You will have "m" in two places in your equation. Use algebra to solve for m.
I repeat: What is the net force being exerted upon the fork? 


#9
Mar511, 08:15 PM

P: 14

I can get up untill
Fapp  m a(gravity) / a(applied) = m Now I am a bit confused on my next move. Is this right so far? Since m on the left is being multiplied by the acceleration of gravity I think I should divide to get rid of it. But once I do the right side would cancel out to zero. 


#10
Mar511, 08:23 PM

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F_{app}  m g = m a
0.35  m g = m (0.15) Not sure what you use for g: 10 m/s^{2} or 9.8 m/s^{2} or 9.81 m/s^{2} Put in the appropriate number for g & solve for m. 


#11
Mar511, 08:34 PM

P: 14

Yes, but as I said earlier I can get up until
0.35  m g / 0.15 = m I can't successfully eliminate the LH m. I tried dividing and adding it to the RH 


#12
Mar511, 09:39 PM

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Before you divide by 'a', your correct equation, per post 5 2nd line, was
0.35  (m 9.8 m/s2) = m 0.15ms2 leaving off units, then 0.35  (9.8m) = 0.15m Now this is algebra, add 9.8m to both sides 0.35 = 9.95m Now solve for m and convert to grams. 


#13
Mar511, 09:54 PM

P: 14

Thanks, that makes allot of sense.



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