Slight Confusion in Introductory Mechanics (Dynamics Problem)

[Lunar Manatee]

Homework Statement

A force F acts on a body of mass 1kg which moves from rest in a straight line starting from the origin point "O". If F=5x+6 where x is its displacement from the origin measured in metre, F in newton, then its velocity where r=4m (editor's note: that's the position vector) is ... m/sec

Relevant Equations

F=ma
a=vdv/dx

the attempt is visible in the image. my question is more... pedantic (?). if the initial conditions are to be taken into account, v_0=0, x_0=0 and F_0=+6, then isn't the velocity supposed to be v=+8sqrt(2)? or am I grasping at straws here.

my thought is, if the force is acting in the positive direction, then the acceleration is also acting in that positive direction, which means the velocity increases in that positive direction too (from 0) which should mean it's only larger than or equal to 0 on the interval t belongs to [0,+inf[

Model Answer says it's (c). I am just having trouble supporting either (c) or (a) as a valid answer.

Edit (1): I think I just realised what the flaw in my logic was; I appreciate the replies if they do come in! I hope they help cement more less-flawed logic in my brain for later use.

 

[PeroK]

IMO, the question is a mess. Vectors and the magnitude of Vectors are horribly mixed up.

 

[Lunar Manatee]

IMO, the question is a mess. Vectors and the magnitude of Vectors are horribly mixed up.

I am guessing they meant to say it was the algebraic (is that how you spell it?) measures. just the component of the vector along the straight-line

 

[PeroK]

I am guessing they meant to say it was the algebraic (is that how you spell it?) measures. just the component of the vector along the straight-line

None of the answers is a vector quantity.

 

[kuruman]

Your reasoning is correct. The force does not change direction with position and, since the object starts from rest, the velocity is always in the same (positive) direction. There is only one choice that is positive.

A negative might be possible if the object had initial velocity and the acceleration were in opposite directions. For example, when you throw a rock straight up in the air, its velocity will be double-valued at heights between the launching point and maximum height.

Although saying if $v^2=\sqrt{128}$ implies $v=\pm 8\sqrt{2}$ is mathematically correct, this mathematical formulation does not describe the physical reality of the question. The person who claimed that the answer is (c) did the math correctly, but then did not exclude the negative answer as unphysical.

 

[kuruman]

None of the answers is a vector quantity.

This is a one-dimensional problem. I think is it safe to assume the standard convention "to the right is positive and to the left is negative."

 

[Steve4Physics]

@Lunar Manatee, note that the SI symbol for the second is 's' not 'sec', so 'm/sec' is incorrect.

EDITed to remove surplus information.

 

[Lunar Manatee]

@Lunar Manatee, note that the SI symbol for the second is 's' not 'sec', so 'm/sec' is incorrect.

EDITed to remove surplus information.

yeah, I have also been arguing that for a while. but it seems *their* convention was: grams (gm), seconds (sec.) and a few other messed up units.

 

[PeroK]

This is a one-dimensional problem. I think is it safe to assume the standard convention "to the right is positive and to the left is negative."

Even so, IMO this is the sort of problem that undermines the student's understanding of physics. And promotes instead a weird, non-physical mathematics.

 

[PeroK]

Your reasoning is correct. The force does not change direction with position and, since the object starts from rest, the velocity is always in the same (positive) direction. There is only one choice that is positive.

A negative might be possible if the object had initial velocity and the acceleration were in opposite directions. For example, when you throw a rock straight up in the air, its velocity will be double-valued at heights between the launching point and maximum height.

Although saying if $v^2=\sqrt{128}$ implies $v=\pm 8\sqrt{2}$ is mathematically correct, this mathematical formulation does not describe the physical reality of the question. The person who claimed that the answer is (c) did the math correctly, but then did not exclude the negative answer as unphysical.

The same constraints exist in the mathematics. So, the mathematics is also wrong.

 

[jbriggs444]

A negative might be possible if the object had initial velocity and the acceleration were in opposite directions. For example, when you throw a rock straight up in the air, its velocity will be double-valued at heights between the launching point and maximum height.

The acceleration need not and does not switch signs.

If we extended the scenario continuously into the past, we would see that the object did indeed have a negative velocity in the past. It arrived from the right with a negative velocity and departed to the right with a positive velocity. Its acceleration never changed sign.

During this extended scenario there were two instants when the object was at $r=4\text{m}$. The example given by @kuruman of a rock thrown straight up is entirely apt.

The key difficulty with the problem is that we when we transcribed the physical situation into mathematics, we did not transcribe our implicit assumption that the scenario excluded times prior to the starting condition.

If we'd transcribed "subject to the constraint that $t \ge 0$" then the mathematics could have eliminated the possibility that $v(t)$ could be negative.

 

[PhDeezNutz]

OP are you familiar with solving second order linear non-homogenous differential equations?

I was able to get the correct answer (Which is A btw as others have pointed out)

Use the ansatz

$x(t) = Ae^{\sqrt{5} t} + Be^{-\sqrt{5} t} + C$

Apply the initial conditions $x(0) = 0$ and $\frac{dx}{dt} (0) = 0$ and finding constants $A,B,C$ should be straightforward

Use some hypertrig definitions for hyperbolic sine, hyperbolic cosine, and a Pythagorean identity

And $8\sqrt{2}$ should pop out no problem

 

[Lunar Manatee]

OP are you familiar with solving second order linear non-homogenous differential equations?

I was able to get the correct answer (Which is A btw as others have pointed out)

Use the ansatz

$x(t) = Ae^{\sqrt{5} t} + Be^{-\sqrt{5} t} + C$

Apply the initial conditions $x(0) = 0$ and $\frac{dx}{dt} (0) = 0$ and finding constants $A,B,C$ should be straightforward

Use some hypertrig definitions for hyperbolic sine, hyperbolic cosine, and a Pythagorean identity

And $8\sqrt{2}$ should pop out no problem

I am familiar with linear non-homogenous differential equations, just didn't really have the wherewithal to confirm it that way; but I am extremely glad you did it and confirmed it, gives a new insight!

 

[kuruman]

The acceleration need not and does not switch signs.

I never said that the acceleration changes signs. There is ambiguity as to the object's state of motion for $t<0.$ The statement of the problem tells us that the object "moves from rest in a straight line starting from the origin point O . . . " The obvious interpretation is that the object is at rest at $t<0$ and starts moving when the force is "turned on" at $t=0.$ A less obvious interpretation is that object is already moving at $t<0$ and comes to rest instantaneously at $t=0.$

Since the force is conservative, this problem can also be solved by using mechanical energy conservation. Taking the zero of potential energy at the origin, the potential energy function is $$U(x)=-\frac{5}{2}x^2-6x.$$ If it is at rest at the origin, it will move in the direction of decreasing potential energy, i.e. +x and keep moving in that direction.

A plot of $U(x)$ is shown on the right. There is a maximum at $x=-1.2~\text{m}$ where the force is zero. It the object is on the negative axis moving towards the origin, it can (a) go over the hump; (b) stay at the hump in unstable equilibrium; (c) never reach the hump and go back where it came from. None of these possibilities allows for the object to be at the origin and have zero velocity.

Therefore, if the object is moving at $t<0$, it must have negative velocity such that it reaches the origin with zero speed.

We see, then, that at $x=4~\text{m}$, the

• obvious interpretation (the object is at rest at the origin until the force is turned on) leads to the velocity $v=+8\sqrt{2}~\text{m/s}.$
• less obvious interpretation (the object is already moving to the left so that it reaches the origin with zero speed) leads to the velocity $v=-8\sqrt{2}~\text{m/s}.$

If I were asking this, I would make it a multiple answer question in which (a) and (b) would be correct and make it clear that we know that the object has zero velocity at O but we don't know whether it is an instantaneous value or not.

 

[Lunar Manatee]

I never said that the acceleration changes signs. There is ambiguity as to the object's state of motion for $t<0.$ The statement of the problem tells us that the object "moves from rest in a straight line starting from the origin point O . . . " The obvious interpretation is that the object is at rest at $t<0$ and starts moving when the force is "turned on" at $t=0.$ A less obvious interpretation is that object is already moving at $t<0$ and comes to rest instantaneously at $t=0.$

View attachment 363025Since the force is conservative, this problem can also be solved by using mechanical energy conservation. Taking the zero of potential energy at the origin, the potential energy function is $$U(x)=-\frac{5}{2}x^2-6x.$$ If it is at rest at the origin, it will move in the direction of decreasing potential energy, i.e. +x and keep moving in that direction.

A plot of $U(x)$ is shown on the right. There is a maximum at $x=-1.2~\text{m}$ where the force is zero. It the object is on the negative axis moving towards the origin, it can (a) go over the hump; (b) stay at the hump in unstable equilibrium; (c) never reach the hump and go back where it came from. None of these possibilities allows for the object to be at the origin and have zero velocity.

Therefore, if the object is moving at $t<0$, it must have negative velocity such that it reaches the origin with zero speed.

We see, then, that at $x=4~\text{m}$, the

• obvious interpretation (the object is at rest at the origin until the force is turned on) leads to the velocity $v=+8\sqrt{2}~\text{m/s}.$
• less obvious interpretation (the object is already moving to the left so that it reaches the origin with zero speed) leads to the velocity $v=-8\sqrt{2}~\text{m/s}.$

If I were asking this, I would make it a multiple answer question in which (a) and (b) would be correct and make it clear that we know that the object has zero velocity at O but we don't know whether it is an instantaneous value or not.

I am thoroughly impressed! I had never thought of analysing it in such a way!

 

[kuruman]

I am thoroughly impressed! I had never thought of analysing it in such a way!

As you do more physics, you will eventually come to the realisation that mechanical energy conservation is a good place to start when you are asked to relate speed and position at point A with speed and position at point B because that's exactly what mechanical energy conservation allows you to do.

(Edited for typos)