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January 2nd, 2019, 07:22 PM   #1
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How to know if an object is at rest and its position given a force versus position di

How to know if an object is at rest and its position given a force versus position diagram?

The problem is as follows:
The force graphed in Figure 1. is applied to a $\textrm{2.0-kg}$ block that was sliding to the right (the $\textrm{+-direction}$) over a frictionless surface with speed $5.0\,\frac{m}{s}$ at $x=0$ (a) Is the block ever at rest? If so,where? (b) Find a position (other than $x=0$) when the block is again moving to the right at $5.0\,\frac{m}{s}$.


The figure alluded in the problem is shown above.

Now the thing with this problem is I really don't know where to begin.

Hence my question is how do I know that the block is at rest?.

My first guess is that this happens when the net force on the block is $0$ or that it is not carrying (unsure if this term is right) kinetic energy?. However can this happen in a frictionless surface?. My other question arises from the fact, how do I understand the negative force in the diagram?. Does it mean that is a force that is being applied to the left side of the block?.

So far the only thing I could come up with wad to use Newton's second law:

$f=ma$

Since what it is being asked is the position, I thought to use:

$v^{2}_f=v^{2}_{0}+2a\Delta x$

For $v^{2}_f=0$ and $v_{0}=5\,\frac{m}{s}$,

$0=\left(5\right)^{2}_{0}+2a\Delta x$

But this would mean to use a, from where? From the diagram?.

If so, at $-40 N$ :

$a=\frac{-40}{2}=-20\,\frac{m}{s^2}$

Then:

$0=\left(5\right)^{2}+2\left(-20\right)\Delta x$

Hence:

$\Delta x=\frac{25}{40}=0.625$

Which would translate as a position $x=0$

$x_{1}-x_{0}=0.625-0=0.625\,m$

So this would be on the right?. Although in the diagram it seems to indicate that the "domain" is in the positive numbers, it doesn't check with the answer. Am I understanding this mathematically or physically wrong?. Can somebody help me to clear these doubts?.
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January 2nd, 2019, 07:58 PM   #2
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$v(t) = \displaystyle \dfrac 1 m\int \limits_0^t~f(\tau)~d\tau + v_0$

$f(\tau)$ is easy enough to discern from the graph

you're given $v_0$

solve for $t$ such that $v(t)=0$

I show there are two such points in time.

$t=\dfrac 5 2,~\dfrac{1}{3} \left(4-\sqrt{10}\right)$

The second question is identical but now solve for $t$ such that $v(t)=5$
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January 3rd, 2019, 08:39 AM   #3
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Quote:
Originally Posted by romsek View Post
$v(t) = \displaystyle \dfrac 1 m\int \limits_0^t~f(\tau)~d\tau + v_0$

$f(\tau)$ is easy enough to discern from the graph

you're given $v_0$

solve for $t$ such that $v(t)=0$

I show there are two such points in time.

$t=\dfrac 5 2,~\dfrac{1}{3} \left(4-\sqrt{10}\right)$

The second question is identical but now solve for $t$ such that $v(t)=5$
Yay! I was hoping an answer which wouldn't use calculus as this problem was obtained from an algebra physics book. Anyways, I'm still confused. Since none of the provided answers do check with what the answers sheet states.
Answer for part a is $1.0\,m$ and for part b is $2.917\,m$.

Again, I'm not very fond with calculus but,

which would be the $f(\tau)$ you're alluding to?

Perhaps you would this? $f(\tau)=30\tau-40$ in $[0,2]$

and also $f(\tau)=20$ in $[2,3]$

and $f(\tau)=20\tau-40$ in $[3,5]$


Since you say to solve for $v(t)=0$,

then,

$v(t) = \displaystyle \dfrac 1 m\int \limits_0^t~f(\tau)~d\tau + v_0$

$m=\,2.0-kg$

$5.0\,\frac{m}{s}$ For brevity purposes I will omit the use of units but they are stated accordingly to what they should be.

$v(t) = \displaystyle \dfrac 1 2\int \limits_0^t~\left(30\tau-40\right)~d\tau + 5$

$v(t) =5 \int \limits_0^t~\left(3\tau-4\right)~d\tau + 5$

$v(t) =5 \left(3[\frac{\tau^2}{2}]^{t}_{0}-4[\tau]^{t}_{0}\right)+5=0$

$v(t)=5 \left(3[\frac{t^2}{2}-0]-4[t-0]\right)+5=0$

Simplifying $5$ and multiplying by $2$ to cancel the denominator at the first term which accompanies $3$.

$v(t)= 3t^{2}-8t+2=0$

Then all seems to solve for both $t$:

$t_{1,2}=\frac{8\pm\sqrt{8^2-4\times 3\times 2}}{2\times 3}$

$t_{1,2}=\frac{8\pm\sqrt{64-24}}{2\times 3}$

$t_{1,2}=\frac{8\pm\sqrt{40}}{2\times 3}$

$t_{1,2}=\frac{4\pm \sqrt{10}}{3}$

$t_{1,2}=\frac{4 \pm \sqrt{10}}{3}$

$t_{1}=\frac{1}{3}\left(4+\sqrt{10}\right)$

or $t_{2}=\frac{1}{3}\left(4-\sqrt{10}\right)$

Yet I don't see this very close to what it was proposed as an answer. And more importantly. I can't seem to reach to what you found at least for one of the roots. Can you please revisit with more details?
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January 3rd, 2019, 08:50 AM   #4
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Oh bloody hell... the graph of the force is against position, not time.

I'll have to rework this. Stay tuned.
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January 3rd, 2019, 09:14 AM   #5
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Quote:
Originally Posted by Chemist116 View Post
How to know if an object is at rest and its position given a force versus position diagram?

The problem is as follows:
The force graphed in Figure 1. is applied to a $\textrm{2.0-kg}$ block that was sliding to the right (the $\textrm{+-direction}$) over a frictionless surface with speed $5.0\,\frac{m}{s}$ at $x=0$ (a) Is the block ever at rest? If so,where? (b) Find a position (other than $x=0$) when the block is again moving to the right at $5.0\,\frac{m}{s}$.
net work = change in kinetic energy

$W_{net} = \dfrac{1}{2}m(v_f^2 - v_0^2)$

$\displaystyle \int_{x_0}^{x_f} F(x) \, dx = \dfrac{1}{2} \cdot (2)(0 - 5^2)$

for $0 \le x \le 2$, $F(x) = 30x - 40$

$\displaystyle \int_0^{x_f} 30x-40 \, dx = -25$

$\bigg[15x^2-40x \bigg]_0^{x^f} = -25$

$15x_f^2 - 40x_f + 25 = 0$

$3x_f^2 - 8x_f + 5 = 0$

$(3x_f - 5)(x_f - 1) = 0 \implies x_f = \dfrac{5}{3} \text{ or } x_f = 1$


Interpretation of motion ...

Over the position interval $0 \le x < 1$, the block moves right.

At $x=1$, the block comes to rest.

Over the interval $1 < x < \dfrac{5}{3}$, the block moves left.

At $x=\dfrac{5}{3}$, the block comes to rest.

For $x > \dfrac{5}{3}$, the block moves right again.
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Last edited by skeeter; January 3rd, 2019 at 09:43 AM.
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January 3rd, 2019, 11:53 AM   #6
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Quote:
Originally Posted by skeeter View Post
net work = change in kinetic energy

$W_{net} = \dfrac{1}{2}m(v_f^2 - v_0^2)$

$\displaystyle \int_{x_0}^{x_f} F(x) \, dx = \dfrac{1}{2} \cdot (2)(0 - 5^2)$

for $0 \le x \le 2$, $F(x) = 30x - 40$

$\displaystyle \int_0^{x_f} 30x-40 \, dx = -25$

$\bigg[15x^2-40x \bigg]_0^{x^f} = -25$

$15x_f^2 - 40x_f + 25 = 0$

$3x_f^2 - 8x_f + 5 = 0$

$(3x_f - 5)(x_f - 1) = 0 \implies x_f = \dfrac{5}{3} \text{ or } x_f = 1$


Interpretation of motion ...

Over the position interval $0 \le x < 1$, the block moves right.

At $x=1$, the block comes to rest.

Over the interval $1 < x < \dfrac{5}{3}$, the block moves left.

At $x=\dfrac{5}{3}$, the block comes to rest.

For $x > \dfrac{5}{3}$, the block moves right again.
This looks a bit more clearer but only one of the answers check being $\textrm{1 m}$.

Now the interpretation of motion is still the part where I'm confused at, why in $0 \le x < 1$ the block is moving to the right?, would'nt be contrary?. (Although the question implies that is moving to the right).

The answer for part (b) is $\textrm{2.917 m}$ which doesn't seem close to what it was obtained from the equation. Could it be that the book is wrong?. I'm reviewing and checking many times and can't get to that number by following your steps. Help.
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January 3rd, 2019, 12:30 PM   #7
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Quote:
Now the interpretation of motion is still the part where I'm confused at, why in $0≤ x< 1$ the block is moving to the right?, would'nt be contrary?.
The block starts at $v_0 = 5 \, m/s$ in the positive direction with a variable negative acceleration ... the block slows to a stop at $x=1$. This is much like throwing a ball upward ... it has a decreasing positive velocity because its acceleration is negative. When it reaches the top of its trajectory, its velocity = 0.

Over the interval $1 < x < \dfrac{4}{3}$, the block speeds up in the negative direction since velocity and acceleration are both negative.

Over the interval $\dfrac{4}{3} < x < \dfrac{5}{3}$, the block moves in a negative direction, but slows down because acceleration is positive. The block returns to a stop at $x = \dfrac{5}{3}$

For $x > \dfrac{5}{3}$, the block returns to moving in a positive direction because acceleration is positive.


I need to ask ... have you taken a beginning course in mechanics? FYI, one of the biggest misconceptions about motion is that direction of motion is always the same as direction of acceleration. Acceleration is a change in velocity, not velocity itself.
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January 3rd, 2019, 01:19 PM   #8
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as shown previously, velocity as a function of position ...

$v(1) = 0$

$v(5/3) = 0$


$\displaystyle [v(2)]^2 - [v(5/3)]^2 = \int_{5/3}^2 30x-40 \, dx = 5 \implies [v(2)]^2 - 0 = 5 \implies v(2) = \sqrt{5}$


note $F(x) = 20$ on the position interval $2 < x < 3$


$\displaystyle \int_2^3 20 \, dx = 20 = [v(3)]^2 - [\sqrt{5}]^2 \implies v(3) = 5$

The block is again at v = 5 at x = 3.

Using geometry, the area below the x-axis between the force graph and the x-axis from x = 0 to x = 4/3 represents -80/3 J change in energy.

Using geometry, the area above the x-axis between the force graph and the x-axis from x = 4/3 to x = 3 represents +80/3 J change in energy.

Overall net change in energy from x = 0 to x = 3 is 0 J.
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January 3rd, 2019, 09:58 PM   #9
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Quote:
Originally Posted by skeeter View Post
The block starts at $v_0 = 5 \, m/s$ in the positive direction with a variable negative acceleration ... the block slows to a stop at $x=1$. This is much like throwing a ball upward ... it has a decreasing positive velocity because its acceleration is negative. When it reaches the top of its trajectory, its velocity = 0.

Over the interval $1 < x < \dfrac{4}{3}$, the block speeds up in the negative direction since velocity and acceleration are both negative.

Over the interval $\dfrac{4}{3} < x < \dfrac{5}{3}$, the block moves in a negative direction, but slows down because acceleration is positive. The block returns to a stop at $x = \dfrac{5}{3}$

For $x > \dfrac{5}{3}$, the block returns to moving in a positive direction because acceleration is positive.


I need to ask ... have you taken a beginning course in mechanics? FYI, one of the biggest misconceptions about motion is that direction of motion is always the same as direction of acceleration. Acceleration is a change in velocity, not velocity itself.
Exactly this is the part where I'm confused at. How does a block which starts in a positive direction and with negative acceleration (since a variable negative force is indicated in the graph) moves to the right and then with the same sign of acceleration (negative acceleration) moves to the left and with another change of sign in acceleration is moving to the right?. By reading at your words I'll explain what I understood.

Before of that I must say that I can understand the math but visualizing what is happening to the object would be much clearer to me since I'm not good at picturing things in my head. I'm only imagining that the object is moving to the right but slowing down then it stops, then it moves to the left (but how with negative acceleration?) maybe the only explanation is that since is moving to -x the negative acceleration is increasing its speed. (This corresponds much to the analogy you gave with throwing a ball upward), but how to explain what it is happening next, I mean the object moving again to the right?. Could it be a bouncing back?, so the object suddenly stops and then changes direction?. Maybe an animation would help this.

I don't want to keep nagging about calculus use (however I would have preferred a more precalc solution and relying in algebra, whatever). But in one way or another (using integrals or just solving areas for a trapezoid) renders the same result $x=1$ and $x=1+\frac{2}{3}$.

The intervals you mentioned in the discussion on what's happening help, but not much. Does it means that in the interval $1 < x < \dfrac{4}{3}$ the block moves to the left the same as $1 < x < \dfrac{5}{3}$ but because in the first interval the acceleration is negative so as the velocity it is speeding up and in the second interval is slowing down due to positive acceleration and negative velocity?. (This latter part makes much sense to me though). So that it finally stops at $\frac{5}{3}$ to change direction again and move to the right with positive acceleration.

If I wasn't clear, maybe this can summarize what I understood.

$\left [0,1 \right \rangle$ Moves right but slows down, until stops
$\left\langle 1,\frac{4}{3}\right\rangle$Changes direction (left) and speeds up
$\left\langle\frac{4}{3},\frac{5}{3}\right\rangle$ Keeps moving left but slowing down until stops.
$\left\langle\frac{5}{3},5\right\rangle$Changes direction to the right and speeds up.

The above paragraph is what I understood from reading at your passage several times. Again, I'm slow at processing these things in my mind. Let me know if I understood you correctly.

Actually I did learned some of basics in physics years ago but forgot almost everything. At the moment I'm self learning mechanics at a slow pace. My current reference (Rex & Wolfson's Essential College Physics) doesn't go into much details as you explained. Because of this I was totally unaware if I did incurred in the transgression that the direction of motion is the same as of acceleration. In my previous paragraph I explained what I could understand and avoided any implying of such but if it gives that impression I want to stress that my intended words were that motion is related with the sign of the position which correspond with that of the velocity (hence is a vector).

Quote:
Originally Posted by skeeter View Post
as shown previously, velocity as a function of position ...

$v(1) = 0$

$v(5/3) = 0$


$\displaystyle [v(2)]^2 - [v(5/3)]^2 = \int_{5/3}^2 30x-40 \, dx = 5 \implies [v(2)]^2 - 0 = 5 \implies v(2) = \sqrt{5}$


note $F(x) = 20$ on the position interval $2 < x < 3$


$\displaystyle \int_2^3 20 \, dx = 20 = [v(3)]^2 - [\sqrt{5}]^2 \implies v(3) = 5$

The block is again at v = 5 at x = 3.

Using geometry, the area below the x-axis between the force graph and the x-axis from x = 0 to x = 4/3 represents -80/3 J change in energy.

Using geometry, the area above the x-axis between the force graph and the x-axis from x = 4/3 to x = 3 represents +80/3 J change in energy.

Overall net change in energy from x = 0 to x = 3 is 0 J.
I'm a bit confused at the way how to understand this:

$\displaystyle [v(2)]^2 - [v(5/3)]^2 = \int_{5/3}^2 30x-40 \, dx = 5 \implies [v(2)]^2 - 0 = 5 \implies v(2) = \sqrt{5}$

Why did we want to subtract $[v(2)]^2 - [v(5/3)]^2$ and why is it equal to $\int_{5/3}^2 30x-40 \, dx$?

I can relate that integral coming from Force vs position diagram (hence Work) but here why is it related with the squares of both velocities? Do you mean due to the difference in its kinetic energies, but this would require to multiply this to the mass or $2-kg$ isn't it?. Which would make $\frac{1}{2}\times\,\textrm{2 kg}\left(v(2)^2-v(5/3)^2\right)$. Am I understanding this part from your explanation correctly?

This part I got stuck at and as well on how do we conclude this.

The block is again at v = 5 at x = 3.

Rereading your explanation seems that by already obtaining $v(2)=\sqrt{5}$ we can find what is going to be the speed at $v(3)$ (using the same difference of kinetic energies as work principle referred in earlier lines) which in the end turns out to be $v(3)=5$. I must confess that I had to revisit this as it was not that obvious because it seemed that the $2-kg$ mass was omitted due being canceled out by the formula. But I ponder, what was the reason of doing all of this? To demonstrate that $x=3$ the speed is again $5$ on what purpose? Is it related with what is it being asked, in other words that the block is moving to the right? (Although it was expressed earlier that this was happening passed $x=1+\frac{2}{3}$.) This part is where I'm still confused.

The latter discussion on the geometry helps, and I thank about that. Needless to say that the key into solving this problem was to relate the work that is needed to counter the initial made by the block at $v_{0}=5\,\frac{m}{s}$. However, I'm still dumbfounded on why the second answer doesn't check!?

As mentioned above, the logic seems right and so on. But where does that number in the answer comes from? I mean $2.917\,\textrm{m}$ to which is alluded to be the other position when the block moves again to the right. Although this number belongs to the interval $x>\frac{5}{3}$ it seems to imply that it was $x>2.917$ which is don't know where did it came from?. Or was just an error in the book? Does it exist a way to "force" and obtain this result?

The doubts I have first is what you tried to say with the velocity as a function of position and the second part, did the book was wrong with that number?

Last edited by skipjack; January 4th, 2019 at 01:03 AM. Reason: added insights
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January 4th, 2019, 03:01 AM   #10
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It seems that skeeter obtained $15x_f^2 - 40x_f + 25 = v_f^2$, the left-hand side of which would need to be non-negative (as the right-hand side is non-negative).

However, the left-hand side is negative for $1 < x< 5/3$.
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