Solving $\vec F = m\vec a$: How is $\vec F = -mkv$ Possible?

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

My working is

$\vec F = m\vec a$
$\vec F = \frac {m(v - v_i)}{\Delta t}$
$\vec F = \frac {-mkx}{\Delta t}$

However, the solution said $\vec F = -mkv$. I don't understand how this is possible since $v ≠\frac {x}{\Delta t}$ since $v =\frac {\Delta x}{\Delta t}$?

Many thanks!

 

[Frabjous]

Homework Statement:: Please see below
Relevant Equations:: Please see below

For this problem,
View attachment 322236
My working is

$\vec F = m\vec a$
$\vec F = \frac {m(v - v_i)}{\Delta t}$
$\vec F = \frac {-mkx}{\Delta t}$

However, the solution said $\vec F = -mkv$. I don't understand how this is possible since $v ≠\frac {x}{\Delta t}$ since $v =\frac {\Delta x}{\Delta t}$?

Many thanks!

You do not take a finite difference, you take the derivative of v with respect to t to get acceleration.

 

[ChiralSuperfields]

You do not divide by δt, you take the derivative of v with respect to t to get acceleration.

Thank you for your reply @Frabjous!

I was trying to solve this problem using the algebra definition of velocity only. It's possible to solve this problem without calculus right?

Many thanks!

 

[Frabjous]

It's possible to solve this problem without calculus right?

Yes, but you took the finite difference wrong. You want start with Δv/Δt.
Hint: Δv_i=0 since it is a constant.

 

[ChiralSuperfields]

Yes, but you took the finite difference wrong. You want start with Δv/Δt.
Hint: Δv_i=0 since it is a constant.

Thank you for your reply @Frabjous !

I though $\Delta v = v_f - v_i = v - v_i$ since they drop the $f$ for $v$

Many thanks!

 

[Frabjous]

Thank you for your reply @Frabjous !

I though $\Delta v = v_f - v_i = v - v_i$ since they drop the $f$ for $v$

Many thanks!

v_i is just a constant. It could have just as easily been labeled a. v_f is not in the original problem.
Δv=Δv_i-Δ(kx)
Since k is also a constant you can pull it outside the Δ.

 

[kuruman]

I wouldn't try it without calculus. Calculus is your friend here.
Might this be of help?$$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}.$$

 

[haruspex]

I was trying to solve this problem using the algebra definition of velocity

The "algebra definition" $\frac{\Delta x}{\Delta t}$ is for the average velocity over the time interval $\Delta t$. You need to use the definitions of instantaneous velocity and instantaneous acceleration, which can only be expressed in calculus.

 

[ChiralSuperfields]

Thank you for your replies @Frabjous, @kuruman and @haruspex !

I have figured out how solve to the problem using algebra. The calculus way as a lot faster thought!

$v_i = v(x_i) = v_i - kx_i$
$v_f = v(x_f) = v_i - kx_f$
$\Delta v = v_f - v_i= v_i - kx_f - (v_i - kx_i) = -kx_f + kx_i = k(x_i - x_f) = -k\Delta x$

Then plug $\Delta v$ into Newton II to get their answer :)

 

[haruspex]

Thank you for your replies @Frabjous, @kuruman and @haruspex !

I have figured out how solve to the problem using algebra. The calculus way as a lot faster thought!

$v_i = v(x_i) = v_i - kx_i$
$v_f = v(x_f) = v_i - kx_f$
$\Delta v = v_f - v_i= v_i - kx_f - (v_i - kx_i) = -kx_f + kx_i = k(x_i - x_f) = -k\Delta x$

Then plug $\Delta v$ into Newton II to get their answer :)

I assume your final steps are
$\frac{ \Delta v }{\Delta t}= -k\frac{ \Delta x }{\Delta t}$
$a=-kv$
But as I pointed out, that doesn’t quite work because the final line above should read
$a_{avg}=-kv_{avg}$, where the averages are taken over the time interval.
To get the target answer you need to apply the calculus principle of letting the deltas tend to zero.

 

[ChiralSuperfields]

I assume your final steps are
$\frac{ \Delta v }{\Delta t}= -k\frac{ \Delta x }{\Delta t}$
$a=-kv$
But as I pointed out, that doesn’t quite work because the final line above should read
$a_{avg}=-kv_{avg}$, where the averages are taken over the time interval.
To get the target answer you need to apply the calculus principle of letting the deltas tend to zero.

Thank you for your reply @haruspex!

Your assumption of my finial steps is correct

I agree that I should has used average notation including $F_{avg}$

However, would you please know why are looking for the instantaneous force not average force from the question?

Many thanks!

 

[kuruman]

However, would you please know why are looking for the instantaneous force not average force from the question?

My answer would be because Newton's law is not $F_{avg}=ma_{avg}$. Perhaps @haruspex has some other answer.

 

[ChiralSuperfields]

My answer would be because Newton's law is not $F_{avg}=ma_{avg}$. Perhaps @haruspex has some other answer.

Thank you for your reply @kuruman !

 

[haruspex]

why are looking for the instantaneous force not average force

Because the question does not specify average it should be taken as meaning instantaneous.

 

[ChiralSuperfields]

Because the question does not specify average it should be taken as meaning instantaneous.

Thank you for your reply @haruspex !