Solving Box Motion with Applied Force

[uncensored188]

Homework Statement

A 3.10 box is moving to the right with speed 8.50 on a horizontal, frictionless surface. At t=0, a horizontal force is applied to the box. The force is directed to the left and has magnitude F(t)= (6.00 N/s^2)t^2


What distance does the box move from its position at t=0 before its speed is reduced to zero?



If the force continues to be applied, what is the speed of the box at 4.00s ?

Homework Equations

The Attempt at a Solution

 

[SammyS]

Homework Statement

A 3.10 box is moving to the right with speed 8.50 on a horizontal, frictionless surface. At t=0, a horizontal force is applied to the box. The force is directed to the left and has magnitude F(t)= (6.00 N/s^2)t^2

What distance does the box move from its position at t=0 before its speed is reduced to zero?

If the force continues to be applied, what is the speed of the box at 4.00s ?

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

Welcome to PF !

You need to make an attempt at a solution before we can help you. It's in the rules for these forums. It's even true for God. !

What is the 3.1 associated with the box?

It's mass?
It's volume?
It's temperature?​

And what are the units?

Find the acceleration of the box.

How are velocity and acceleration related?

 

[DaveC426913]

You must show your attempts.

[EDIT Ah. Beat me.]

 

[uncensored188]

1.15997\units{m}
3.47\units{m}
11.598\units{m}
10.038\units{m}

all wrong for the first part

 

[basenne]

Show your work. That way we can better assist you.

 

[uncensored188]

its a 3.1 kg box and i have no clue how to approach it

 

[cepheid]

its a 3.1 kg box and i have no clue how to approach it

Basically, because of Newton's 2nd law, you know the acceleration as a function of time a(t). How can you use this to figure out the velocity as a function of time v(t)? Hint: what are the definitions of acceleration and velocity?

 

[uncensored188]

i know you have to integrate from acceleration to velocity but it is too confusing

 

[cepheid]

i know you have to integrate from acceleration to velocity but it is too confusing

Good! Show us the first few steps of the integration that you attempted, up to the point where you got stuck. Explain specifically where you got confused. Then we can help you. We're more than willing to help you learn, but we're not just going to do it for you. Sorry.

 

[uncensored188]

can i use the formula ft=mv

6t^3=3.10 * 8.5

and i get that time equals 1.6376 and and i get acceleration is 6t^2/m

what formula do i plug this into, I'm not too good at integrating I'm completely confused

 

[cepheid]

can i use the formula ft=mv

No, you can't, because this is only true if the force is constant. The more general relation is ∫F(t)dt = Δp (impulse is the integral of force with respect to time, and it is equal to the change in momentum).

However, it is not necessary to use impulse and momentum if you don't want to. You can just do what you said you were going to do, which is to integrate the acceleration function with respect to time in order to get the velocity function. Once you have the velocity as a function of time v(t), it should be easy to solve for the value of t at which v(t) = 0. In order to integrate the acceleration function a(t), you must first figure out what it is, and you have done so:

and i get acceleration is 6t^2/m

Correct! This is a(t). Now integrate it to get v(t)!

 

[uncensored188]

thank you guys very much for your help, I figured it out
after i integrated i used 8.5 for c and and then took the definite integral from 0 to the time i found when velocity is 0. I got the correct answer!

 

[monikraw]

can someone please clarify this to me? I have a problem very similar to this one.

what exactly are you integrating and what do you get once you integrate?

 

[SammyS]

can someone please clarify this to me? I have a problem very similar to this one.

what exactly are you integrating and what do you get once you integrate?

This thread is one year old.

Integrating acceleration gives velocity, i.e, acceleration is the derivative of velocity, so velocity is the anti-derivative of acceleration.