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anonymous
 one year ago
A force that depends on time is given as (5.00i3.00j). It exerts on a 2.00kg 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)
anonymous
 one year ago
A force that depends on time is given as (5.00i3.00j). It exerts on a 2.00kg 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)

This Question is Closed

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\(5.00 \hat i3.00 \hat j\) does not depend on time, the components are constant, it is constant maybe you missed somthing out?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It is.a time dependent force. It can written as this vector notation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Think of it as (5.00i3.00j)t First get acceleration using force/mass. Then set acceleration equals to dv/dt

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1great 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
 one year ago
Best ResponseYou've already chosen the best response.0You want the steps? I have the correct answers and steps if you want to see/

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1with the leaking bucket idea, go for it!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The mass would have to be constant for it to work.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1for newton's law to work?!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You 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
 one year ago
Best ResponseYou've already chosen the best response.1you can solve it \(\vec F = m \frac{d(m\vec v)}{dt}\) which is what i meant by the original formulation then \[<5,3>t = \frac{d [(20.1)t \ \ \vec v ]}{dt}\] in the x direction \[5t = 0.1 v_x + (20.1t) \frac{dv_x}{dt}\] write that as \[ \frac{dv_x}{dt} \frac{0.1}{(20.1t)} v_x = \frac{5t}{(20.1t)} \] and you can solve with an integrating factor same in the y direction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0X and y component should have equal times. This is a single two dimensional force. So you mean (53)t instead?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For first equation I cant read that. Can you write it without the equation sheet?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Its 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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