I don't understand force as a function of time

[fishturtle1]

Homework Statement

A 2.70 kg box is moving to the right with speed 9.50 m/s 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/s2 )t2

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

Homework Equations

F = ma
vf^2 = vi^2 + 2ad
xf = xi + vit + .5axt^2

The Attempt at a Solution

So i am confused about this F(t) = 6.00(n/s2)t^2

I've read solutions to problems similar to this where they say to integrate to get a velocity equation and then solve for v=0.

So I did that:

a=6t^2
v=(6t^3)/3 + C, and plugged in my initial velocity, 9.50 m/s as C. I am not sure why but I saw someone do this in the solutions I looked at.

From this I solved for time and got 1.68s.

Then I used Newton's 2nd Law .

F = ma = 6.00t^2. I didn't include weight and normal force because the object is only moving along the x axis.

ma = 6.00t^2
2.70a=6.00(1.68^2)
a = 6.272m/s^2

Now that I have acceleration, initial velocity, initial position, and time, I use my kinematic equation to find the final position where Vf is 0.

xf = 0 + 9.5(1.68) + .5(6.272)(1.68^2)
xf= 15.96 + 8.85
xf = 24.8m

This is wrong. The parts where I am very confused are interpreting what F(t) = 6.00(N/s2)t^2 means compared to F(t) = 6.00N. I am also confused with knowing how to determine what C is, but maybe I would understand that part if I understood the force function. Thank you for looking at this!

 

[PeroK]

Homework Statement

A 2.70 kg box is moving to the right with speed 9.50 m/s 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/s2 )t2

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

Homework Equations

F = ma
vf^2 = vi^2 + 2ad
xf = xi + vit + .5axt^2

The Attempt at a Solution

So i am confused about this F(t) = 6.00(n/s2)t^2

I've read solutions to problems similar to this where they say to integrate to get a velocity equation and then solve for v=0.

So I did that:

a=6t^2
v=(6t^3)/3 + C, and plugged in my initial velocity, 9.50 m/s as C. I am not sure why but I saw someone do this in the solutions I looked at.

From this I solved for time and got 1.68s.

Then I used Newton's 2nd Law .

F = ma = 6.00t^2. I didn't include weight and normal force because the object is only moving along the x axis.

ma = 6.00t^2
2.70a=6.00(1.68^2)
a = 6.272m/s^2

Now that I have acceleration, initial velocity, initial position, and time, I use my kinematic equation to find the final position where Vf is 0.

xf = 0 + 9.5(1.68) + .5(6.272)(1.68^2)
xf= 15.96 + 8.85
xf = 24.8m

This is wrong. The parts where I am very confused are interpreting what F(t) = 6.00(N/s2)t^2 means compared to F(t) = 6.00N. I am also confused with knowing how to determine what C is, but maybe I would understand that part if I understood the force function. Thank you for looking at this!

Your kinematics equations only hold for constant acceleration. In this case, you have a variable acceleration that is changing with time.

You need, therefore, to apply calculus to calculate distance and velocity over time. That's where the integration comes in.

Have you covered some course material on this?

PS A force of $F = 6N$ is a constant force. A force of $F = 6t^2 (N/s^2)$ is a force that starts at $0$ and increases with time, according to that formula. At $t=10s$, say, the force would be $600N$. You should be familiar with a function like that from your maths classes.

 

[CWatters]

I've read solutions to problems similar to this where they say to integrate to get a velocity equation and then solve for v=0. So I did that:

a=6t^2

You need to apply Newtons law first eg..

f=ma
so
a=f/m
then replace f with 6t²
a=6t²/m

then integrate to give velocity etc

 

[fishturtle1]

Thank you I used this information to solve both parts of the problem. I use the integral of acceleration and of velocity to find time and then used that time to find position.

 

[UniqueName]

Thank you I used this information to solve both parts of the problem. I use the integral of acceleration and of velocity to find time and then used that time to find position.

How did you do it? I'm having some trouble on this question. This is what I got
a = (6t^2)/5
v = (2t^3)/5
x = (t^4)/10

9.5m/s = ((2N/s^2)t^3)/5kg
47.5N/s = (2N/s^2)t^3
23.75s=t^3
2.87s I'm not sure if the units are right here

then I use that in the x equation

x = (2.87s)^4/10 = 6.83 m? not sure about the units here either.

do I have to use another kinematics equation?

 

[haruspex]

a = (6t^2)/5
v = (2t^3)/5

Include units, watch the signs (which are you taking as positive, left or right?) and don’t forget the constants of integration and initial conditions.

 

[CWatters]

a = (6t^2)/5

Mass is 2.7kg not 5kg? Or is your problem slightly different?

 

[UniqueName]

a = (6t^2)/5

Mass is 2.7kg not 5kg? Or is your problem slightly different?

I had misread it. The numbers are different for everyone, I just realized mine was 1.5kg not 5kg

 

[UniqueName]

Include units, watch the signs (which are you taking as positive, left or right?) and don’t forget the constants of integration and initial conditions.

Like this?
a = ((6 N/s²) t²)/1.5kg
v = ((3 N/s²) t³)/1.5kg
I don't think the units change very much but I could be wrong

 

[haruspex]

Like this?
a = ((6 N/s²) t²)/1.5kg
v = ((3 N/s²) t³)/1.5kg
I don't think the units change very much but I could be wrong

Yes, but I should have also said to work algebraically as far as possible. That means you won't need to worry about units until plugging in numbers at the end.
With regard to getting a solution, my other two recommendations are more relevant.