## anonymous one year ago 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)

1. 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?

2. anonymous

It is.a time dependent force. It can written as this vector notation.

3. anonymous

Think of it as (5.00i-3.00j)t First get acceleration using force/mass. Then set acceleration equals to dv/dt

4. 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!

5. anonymous

You want the steps? I have the correct answers and steps if you want to see/

6. IrishBoy123

with the leaking bucket idea, go for it!

7. anonymous

The mass would have to be constant for it to work.

8. IrishBoy123

for newton's law to work?!

9. 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.

10. 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

11. anonymous

X and y component should have equal times. This is a single two dimensional force. So you mean (5-3)t instead?

12. anonymous

For first equation I cant read that. Can you write it without the equation sheet?

13. anonymous

Its a long procedure. Right now I am too lazy to post the steps. Props for trying and making it more interesting.