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I don't understand force as a function of time

  • #1
264
28

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!
 

Answers and Replies

  • #2
PeroK
Science Advisor
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Insights Author
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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.
 
Last edited:
  • #3
CWatters
Science Advisor
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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
 
  • #4
264
28
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.
 

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