anonymous
  • anonymous
A force that depends on time is given as (5.00i-3.00j). It exerts on a 2.00-kg object initially at rest. What time will the object be moving at a speed of 18.0 m/s? Go try it! (I will post the steps and answer later)
Physics
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SOLVED
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chestercat
  • chestercat
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IrishBoy123
  • IrishBoy123
\(5.00 \hat i-3.00 \hat j\) does not depend on time, the components are constant, it is constant maybe you missed somthing out?
anonymous
  • anonymous
It is.a time dependent force. It can written as this vector notation.
anonymous
  • anonymous
Think of it as (5.00i-3.00j)t First get acceleration using force/mass. Then set acceleration equals to dv/dt

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IrishBoy123
  • IrishBoy123
great idea @Shalante but let's make it more interesting. the mass of the object changes too. so it is like a leaking bucket with \(\frac{dm}{dt} = -0.1\) then we really have to get into Newton's Second law and how it was originally formulated go for it!
anonymous
  • anonymous
You want the steps? I have the correct answers and steps if you want to see/
IrishBoy123
  • IrishBoy123
with the leaking bucket idea, go for it!
anonymous
  • anonymous
The mass would have to be constant for it to work.
IrishBoy123
  • IrishBoy123
for newton's law to work?!
anonymous
  • anonymous
You cant make the mass change or else the acceleration or velocity would be impossible to solve. It would be possible if the force was not time dependent.
IrishBoy123
  • IrishBoy123
you can solve it \(\vec F = m \frac{d(m\vec v)}{dt}\) which is what i meant by the original formulation then \[<5,-3>t = \frac{d [(2-0.1)t \ \ \vec v ]}{dt}\] in the x direction \[5t = -0.1 v_x + (2-0.1t) \frac{dv_x}{dt}\] write that as \[ \frac{dv_x}{dt} -\frac{0.1}{(2-0.1t)} v_x = \frac{5t}{(2-0.1t)} \] and you can solve with an integrating factor same in the y direction
anonymous
  • anonymous
X and y component should have equal times. This is a single two dimensional force. So you mean (5-3)t instead?
anonymous
  • anonymous
For first equation I cant read that. Can you write it without the equation sheet?
anonymous
  • anonymous
Its a long procedure. Right now I am too lazy to post the steps. Props for trying and making it more interesting.

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