# Nonconstant Acceleration Problem

• Hydrinium.h2
In summary: I found out what I was doing wrong, in |Ff|=4.3(1+x0.5)N, N is Newtons not the normal force. If you don't multiply by the normal force, you end up with 2.89=Vf^2 and that gives 1.7 as an answer, which is one of the possible selections.In summary, a 2 kg box is moving at 2 m/s, it is being slowed by friciton such Ff| = 4.3(1+x0.5) N. After moving 20 cm, the speed of the box is 0.58 m/s.
Hydrinium.h2

## Homework Statement

A 2 kg box is moving at 2 m/s, the box is being slowed by friciton such| Ff| = 4.3(1+x0.5) N. What is the speed of the box after moving 20 cm?

## Homework Equations

W=1/2mV^2-1/2mV0^2
W=FΔxCos(θ)
0=N+∑Fy
Friction=μN

## The Attempt at a Solution

So I realize that this is a nonconstant acceleration problem, meaning total work ends up being the integral of FΔxCos(θ), rather than just FΔx.
My knowns are:
Friction=Ff=4.2(1+x^.5)N
Vf= ?
V0= 2m/s
m=2kg
Weight=Fg=2kg*-9.8m/s^2=-19.6N
Normal=N=?
Time=?
ΔX=.2m

I can find normal using 0=N+∑Fy, where all forces pushing in the downward direction are canceled out by the normal force. So N=-∑Fy and since there is only gravity pushing on the box, the Normal force is 19.6 N upwards.

I input the normal force into the friction formula to get:
Ff=4.2(1+x^.5)19.6=84.28+82.28x^.5

Work also equals 1/2mV^2-1/2mV0^2, inputting all I know, I get 1/2*2*V^2-1/2*2*2^2=(V^2)-4

The only force working on the system in the direction of work (x-axis) is friction, so the net force is Ff. Therefore, W=FΔxCos(θ)=FfΔxCos(180)=-FfΔx

I set this equal to (V^2)-4
-FfΔx=(V^2)-4

This is where I begin to struggle. I am pretty sure you need to integrate the left side of the equation, since this problem has nonconstant acceleration, but I'm bad at calculus and also unsure about the physics. Anyways, I ended up with
-FfΔx=-(84.28+84.28x^.5)dx, integrating from 0 to .2
56.187x^1.5+85.28x (from 0 to 0.2)=-21.8815
-21.8815=V^2-4
-17.88=V^2
... and that's where I grind to a halt.
I'm sure I did something completely wrong, or totally interpreted the problem incorrectly but for the life of me I can't figure out what. I would really appreciate some feed back on what I did wrong and what I can do to solve this problem.
Also, this is a multiple choice problem so the choices are:
1.7 m/s
2.66 m/s
0.58 m/s
1.42 m/s
Thanks!

I don't understand why you calculate any normal forces.
The force from friction is given in the problem statement, you don't have to calculate it (especially not with a different result...). You can assume it acts directly against the direction of motion, the problem is 1-dimensional and you don't have to care about angles.

| Ff| = 4.3(1+x0.5) N
Do you mean ##| Ff| = 4.3(1+x^{0.5}) N##?
What is x? The length in some unit?

Work also equals 1/2mV^2-1/2mV0^2, inputting all I know, I get 1/2*2*V^2-1/2*2*2^2=(V^2)-4
Don't forget units.
W=FΔxCos(θ)
This does not work if the force is not constant over Δx.
This is where I begin to struggle. I am pretty sure you need to integrate the left side of the equation
You need an integration, indeed.

-FfΔx=-(84.28+84.28x^.5)dx, integrating from 0 to .2
I guess if you fix the errors I pointed out, the rest should be fine.

Hi, thanks for the help, but I don't understand what errors you pointed out, could you clarify?

1. Yes, that is the equation that I meant. The problem never specified what x was, but I assumed it was the distance traveled at time (t).

2. Thanks, I won't forget units

3. I used the integral of FΔxCos(θ) from point a to b to find the work done by a nonconstant force, as opposed to just FΔx without integration, does this not work? If it doesn't what should I do instead?

4. Thanks for the confirmation

5. I appreciate the help, but could you clarify the errors I made?

I found out what I was doing wrong, in |Ff|=4.3(1+x0.5)N N is Newtons not the normal force. If you don't multiply by the normal force, you end up with 2.89=Vf^2 and that gives 1.7 as an answer, which is one of the possible selections.
Thanks!

Your approach is correct, but there are a few errors in your calculations. First, the normal force should be positive, since it is in the opposite direction of gravity. So N=19.6 N upwards.
Next, when finding the work done by friction, you did not include the cosine of the angle between the force and displacement. In this case, it would be cos(180)= -1. So the work done by friction would be -FfΔx(-1)= FfΔx.
Also, when integrating, you should use the full formula for friction, which is 4.3(1+x^0.5), not just 4.3(1+x^0.5). So the integral should be -4.3(1+x^0.5)dx.
Finally, when plugging in the values, you should use the correct value for Ff, which is 4.3(1+0.2^0.5)= 4.3(1.447)= 6.2121 N.
So the final equation should be -6.2121(0.2)= V^2-4. Solving for V, we get V= 1.7 m/s.
Therefore, the correct answer is 1.7 m/s.

## 1. What is nonconstant acceleration?

Nonconstant acceleration refers to a situation where an object's acceleration is changing over time. This means that the object is not moving at a constant speed, but rather its velocity is either increasing or decreasing.

## 2. What causes nonconstant acceleration?

Nonconstant acceleration can be caused by several factors, such as changes in the forces acting on an object, changes in the object's mass, or changes in the object's direction of motion. Any of these factors can cause an object's acceleration to change.

## 3. How is nonconstant acceleration different from constant acceleration?

Constant acceleration refers to a situation where an object's acceleration remains the same over time. This means that the object is moving at a constant speed, either in a straight line or in a circular motion. Nonconstant acceleration, on the other hand, means that the object's acceleration is changing over time.

## 4. How is nonconstant acceleration calculated?

The calculation for nonconstant acceleration is similar to that of constant acceleration, except that it takes into account the changes in acceleration over time. The formula for nonconstant acceleration is a = (vf - vi) / t, where vf is the final velocity, vi is the initial velocity, and t is the time interval.

## 5. What are some real-life examples of nonconstant acceleration?

Nonconstant acceleration can be seen in many real-life situations, such as a car accelerating from a stop sign, a rollercoaster moving through loops and turns, or a rocket launching into space. In each of these examples, the object's acceleration is changing over time due to external forces acting on it.

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