Dynamic problem

  • Thread starter hastings
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  • #1
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Homework Statement


an object with m=100kg is moving on a straight line. for t=0 its speed is v0=20 m/s. There's a force which is opposing to the movement F=C*v^2 ( C=1kg/m and v is the speed (v(t)=...) ). Find the time t1 in which the speed v=v0/2.


Homework Equations


F=m*a (?)
v(t)=a*t+v0


The Attempt at a Solution


v1=a*t1+v0 --> v0/2=a*t1+v0 --> t1=(-v0/2)*1/a

assuming a (accelation) constant
m*a = F = C*v1^2= C*(v0/2)^2
a= [C*(v0/2)^2]/m
Putting a in
t1= (-v0/2)/a should be getting the answer

Unfortunately the answer I get is not correct (should be 5second)

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
cepheid
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The acceleration is not constant because the force is not constant. Think about it. The drag force is proportional to the square of the velocity.This means that as the object slows down (due to the drag force), the drag force itself decreases. So not only does it decelerate, but it decelerates at an ever decreasing rate. You need a new way to approach this problem. Using the dynamical equations derived for the special case of motion under a constant force doesn't cut it here.
 
  • #3
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then how to solve it?
 
  • #4
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Big hint: Think Newton's Second Law and think calculus....
 
  • #5
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F=m*a=C*v^2
so a=(C*v^2)/m

a=dv/dt ==> v=integral ( (C*v^2)/m )= C*v^2/m * t

so then

t= (v*m)/(C*v^2)=m/(C*v)

I then say
t0=0
t1 -t0=t1
t0=m/(C*v0)
t1= 2m/(C*v0)
t1= t1-t0= m/(C*v0)=5 seconds
 
  • #6
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hastings said:
v=integral ( (C*v^2)/m )= C*v^2/m * t
Umm..not quite.
v is actually dependent on t, and so you can't treat it as a constant.
You should write
[tex]\frac{dv}{dt}=Cv^2/m[/tex]

[tex]\frac{dv}{v^2}=Cdt/m[/tex]
And integrate on both sides.

And don't forget the constant of integration whose value can be determined with the given data.
 
  • #7
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[tex]\int{\frac{1}{v^2}}dv[/tex]? How to do it? I dunno
Maybe [tex]-\frac{1}{v}[/tex]?

Then what about [tex]\int \frac{C}{m}dt= \frac{C}{m} \cdot t+ c_0[/tex]?
 
Last edited:
  • #8
590
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You are perfectly correct, why did the doubts arise?
Anyway can you proceed now?
 
  • #9
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So is it ....

[tex]-\frac{1}{v}+v_0=\frac{C}{m} \cdot t [/tex]

[tex] t=\frac{m}{C} \cdot ( -\frac{1}{v}+v_0)[/tex] ?

Then I substitute [tex]v=\frac{v_0}{2}[/tex] in the last expression and I should get the answer, right?
result should be 5 seconds.
Let's see...
Well...it's not right
 
Last edited:
  • #10
590
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I notice that in the question, it is mentioned that the mass is decelarated. This means you should assign a negative value to acceleration.

Not quite, v_0 refers to velocity at time t=0.

Integrating, you should get
-1/v + K(constant of integration) = -Ct/m

Can you find K?
 
  • #11
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Thought it was K=v0; wouldn't know How can I get it ?
 
  • #12
590
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At t=0, v=20. For LHS=RHS, K= ....
 
  • #13
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For LHS=RHS, K= ....
What's LHS=....? :eek:
Never heard
Is it K=20? but that's what i said previously: K=v0=20
 
  • #14
590
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LHS - Left hand side RHS - Right hand side (for the equation -1/v + K(constant of integration) = -Ct/m) ) :smile:
What I'm saying is put t=0 and v=20 in this equation and find K.
 
  • #15
cepheid
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LHS - left-hand side (of the equation), that's all
 
  • #16
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so k=1/20
[tex]-\frac{1}{v} +\frac{1}{20}=-\frac{C}{m} \cdot t[/tex]

Is it right?
Now I substitute t=t1 and v=v0/2 in the expression?

[tex]-\frac{1}{v} +\frac{1}{20}=-\frac{C}{m} \cdot t1[/tex]

[tex]t1=(-\frac{m}{C}) \cdot (-\frac{1}{v} +\frac{1}{20}) =[/tex]

[tex]=(-\frac{m}{C}) \cdot (-\frac{2}{v_0}+\frac{1}{20})=[/tex]

[tex]=(-\frac{m}{C}) \cdot (\frac{-2 \cdot 20 + v_0}{20 \cdot v_0})=[/tex]

[tex]=(-\frac{m}{C}) \cdot (\frac{-40+v_0}{20 \cdot v_0})=[/tex]

[tex]=(-100) \cdot (\frac{-40+20}{20 \cdot 20})=[/tex]

[tex]=(-100) \cdot (\frac{-20}{400})= 5 \mbox{ seconds }[/tex]

OK! GREAT! Thank you! :blushing: :rofl:
 
Last edited:

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