
#1
Mar912, 01:18 PM

P: 58

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 



#2
Mar912, 01:38 PM

HW Helper
P: 3,394

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. 



#3
Mar912, 01:39 PM

Sci Advisor
HW Helper
Thanks
P: 26,167

hi iJamJL!
and yes the acceleration is the second derviative, even in three dimensions! (what is worrying you about that? ) 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 



#4
Mar912, 01:57 PM

P: 58

Not sure which formula to use! Deals with work and position
Thanks for the replies!
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! 



#6
Mar912, 02:26 PM

P: 58

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. 



#7
Mar912, 03:07 PM

Sci Advisor
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Thanks
P: 26,167





#8
Mar912, 03:18 PM

Mentor
P: 11,445





#9
Mar912, 03:19 PM

P: 58

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? 


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