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)

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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

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

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- 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

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

- IrishBoy123

with the leaking bucket idea, go for it!

- anonymous

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

- IrishBoy123

for newton's law to work?!

- 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

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

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

- anonymous

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

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.