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Particle Kinetics - Impulse

  • Thread starter Doonami
  • Start date
  • #1
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Homework Statement


A 0.2 kg particle moves in the x-y plane under the action of its weight and a force F that varies with time. The velocity of the particle is given by the expression

[tex]\hat{v}[/tex] = 7.5(t2 + 3)[tex]\hat{i}[/tex] - (10/3)(t3 - 4)[tex]\hat{j}[/tex]


Homework Equations


mv1 + [tex]\sum[/tex][tex]\int[/tex]Fdt = mv2


The Attempt at a Solution


Am I crazy? Or Can I just plug in 't=2' into the given equation? I'm sceptical, because it seems too good to be true. It is asking for F at t = 2 and the effect of gravity seems to be accounted for in the vector equation.
 

Answers and Replies

  • #2
76
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hmmm good question
i havent seen this type of question before but if your using the equation you have stated it looks like there should be some limits to deal with the integral.
On second thought, if you integrate F dt, you would end up with Ft which happens to be impulse so to be honest it looks a little confusing.
Maybe plug in your value for t into the velocity equation and then your value for t into the equation you have with the integral and rearrange to find F?
good luck!
 
  • #3
3,763
8
hmmm good question
if you integrate F dt, you would end up with Ft which happens to be impulse so to be honest
You cannot say that because F depends on the variable t !

So integration of Fdt doesn't give you Ft. this would only be the case if F were independent of t.

Now, the OP's question is to calculate F at t=2 no ?

What's the connection between F and v if you know that F=ma (a : acceleration)

marlon
 
  • #4
76
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ahhh nuts this is true!!
sorry to be no help!!
 
  • #5
5
0
a = dv/dt

So, from that I should differentiate the equation for v with respect to time. I can just do this for the individual components (i and j) right? This should yield . . .

ma = 0.2[ (15t)i - (10t2)j ]

multiplied by mass, then sub in t=2.
 
  • #6
Doc Al
Mentor
44,882
1,129
a = dv/dt

So, from that I should differentiate the equation for v with respect to time. I can just do this for the individual components (i and j) right? This should yield . . .

ma = 0.2[ (15t)i - (10t2)j ]

multiplied by mass, then sub in t=2.
Good. This tells you the net force, now find F.
 

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