Not sure which formula to use! Deals with work and position
1. The problem statement, all variables and given/known data
A onedimensional force acts on a particle of mass m = 6.26 kg in such a way that its position is given by: x = 0.484t3  33.6t Find W, the work done by this force during the first 1.49 s. 2. Relevant equations F=mg F=ma W=F*distance 3. The attempt at a solution To find the distance, I plugged in 1.49 into the formula they gave, and I got (48.5). I know that to find work, the formula is W=F*s, so in order to find the force, I'm not sure which formula to use. I think I should be using F=ma, and to find the acceleration, I can take the second derivative of the given formula. However, I forget whether that is allowed in onedimension. If not, then do I use F=mg? If I use F=mg, then the answer for work will come out positive. If I use F=ma, then the work comes out negative. Here are my results: W=Fs W=ma*s W=6.26*18.18*(48.5) W=(5519.6)J W=Fs W=mg*s W=6.26*(9.81)*(48.5) W=2978.4 J 
Re: Not sure which formula to use! Deals with work and position
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Gravity is not involved in this. It would cause a constant acceleration and a distance formula of s = Vi*t + .5*a*t² rather than the cubic formula of this problem. 
hi iJamJL! :smile:
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and yes the acceleration is the second derviative, even in three dimensions! :smile: (what is worrying you about that? :confused:) Quote:
but W = F*s only works if F is constant here, F isn't constant, so you would have to use W = ∫ F.ds however it might be simpler to use the workenergy theorem, and calculate the difference in kinetic energy :wink: 
Re: Not sure which formula to use! Deals with work and position
Thanks for the replies!
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To find the final velocity, which for this case is at 1.49s, we take the first derivative at 1.49: x = 0.484t3  33.6t dx/dt= 2*(.484)t^2  33.6 v= (30.38)m/s W=ΔKE W= 1/2*m(vfinal^2)  1/2*m(vinitial^2) **vinitial is 0 because t=0, x=0** W=1/2*6.26*(30.38)^2 W=2888.8 J I tried entering that into my online homework system and it came out wrong. The system doesn't really care for the form that the answer is, as long as there are 3 significant digits. That means that I solved for this incorrectly! :cry: 
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Re: Not sure which formula to use! Deals with work and position
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My end result came to: v=(.484)*3*(1.49^2)  33.6 =(30.4)m/s W=1/2*m*v^2 W=1/2*6.26*(30.4^2) W=2892.6 J That's still coming out as incorrect. 
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Re: Not sure which formula to use! Deals with work and position
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Re: Not sure which formula to use! Deals with work and position
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v(0)=(33.6)m/s v(1.49)=(30.4)m/s W=1/2*m*(v2^2)  1/2*m*(v1^2) W=1/2*6.26(30.4^2)  1/2*6.26*(33.6^2) W=2892.6  3533.6 W=(641) J Did I finally get the correct answer? :approve: 
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