How long an object takes to stop - using friction, momentum

[gelfand]

Homework Statement

A block of mass 100 grammes was stationary on the flat surface at $x = 0$. At
time $t = 0$ a horizontal force of 10 Newtons was applied on the block in the
positive x direction during $\Delta t = 3$ seconds.

Find at what x position the block will stop, if the coefficient of kinetic
friction between the block and the surface is $\mu_k = 0.2$.

Homework Equations

$I = F \Delta t = m \Delta v = \Delta p$ (where $p$ is used for momentum)

The Attempt at a Solution

I have that

$$
F = ma
$$

Then

$$
\frac{10}{0.1} = a
$$And I can presume (?) that acceleration is constant.Then from this I have

$$
a = 100
$$

So if I have $a = 100$ then I can use Impulse as $I = F \Delta t$ to find

$$
F \Delta t = \Delta p
$$

So

$$
(0.1) (100) (3) = \Delta p = 30
$$So from $\Delta p = 30$ I have $m(v_f - v_0) = 30$ and as $v_0 = 0$ this is
$mv_f = 30$ and $v_f = \frac{30}{0.1} = 300$

This gives me the velocity as $300$ m/s at the point the force has been
removed.

From here I need to know how long it takes for the force of friction to slow the
object to a stop.

This means all of the energy from the point $t = 3$ is spent on friction.

So

$$
\frac{1}{2} mv^2 = \text{work done against friction}
$$So here I have that work done is $W = F \Delta x$ so given $\mu_k = 0.2$, the
mass of the object is $0.1$ then I would have the force of friction as

$$
F_{friction} = mg \times \mu_k = 0.1 \times 9.81 \times 0.2 = 10.11
$$So then I have

$$
\frac{1}{2} (0.1) v^2 = 10.11 \Delta x
$$

And so

$$
\frac{1}{2} (0.1) v^2 \times \frac{1}{10.11} = \Delta x
$$

Using the velocity found from before this gives $$
\frac{1}{2} (0.1) (300)^2 \times \frac{1}{10.11} = \Delta x
$$

So $\Delta x \approx 445.103$ meters.

I don't really have much of a handle on what to expect from an answer here.

I think what I've done is reasonable though?

Thanks

 

[PeroK]

I have that

F=ma​

F = ma

Then

100.1=a​

\frac{10}{0.1} = aAnd I can presume (?) that acceleration is constant.Then from this I have

a=100​

a = 100

You need units here. Otherwise, that looks right.

This gives me the velocity as 300300 m/s at the point the force has been
removed.

Yes, but it would have been much easier just to multiply the acceleration by $3$ seconds!

So here I have that work done is W=FΔxW = F \Delta x so given μk=0.2\mu_k = 0.2, the
mass of the object is 0.10.1 then I would have the force of friction as

Ffriction=mg×μk=0.1×9.81×0.2=10.11​

F_{friction} = mg \times \mu_k = 0.1 \times 9.81 \times 0.2 = 10.11

You should be able to see without a calculator that multiplying $9.81$ by two small fractions is not going to give you something bigger.

Finally, you also need to think about how far the block moves during the acceleration phase.

 

[gelfand]

Ah yes >.< , whoops.

So that the force of friction would be $0.1 \times 9.81 \times 0.2 = 0.1962$.

The means that during the acceleration phase (from $t = 0$ to 3) I have the net
Force would be

$$
F_n = 10 - 0.1962 = 9.8038
$$

Given this I have the work done against friction over the three seconds as

$$
F_n \times \Delta x
$$

But I'm not sure how to find $\Delta x$ from this?

Can I use the conservation of energy here?

From that I would have that $E = U + K + constant$. But I don't really have any
potential energy, and the kinetic energy at the start is zero (so this seems
like a poor choice).

Oh - I can just use the equations of motion here then, so I would have

$$
x = x_0 + v_0 t + \frac{1}{2} a t^2
$$

So here this is just $x = 0.5 \times 100 \times 9 = 450$. So that I would have
moved $350$ meters during $\Delta t$ as a result of the force.

But is that correct? Do these equations work when there's friction present?

 

[PeroK]

So that the force of friction would be 0.1×9.81×0.2=0.19620.1 \times 9.81 \times 0.2 = 0.1962.

Yes, of course, you have friction during the acceleration phase as well.

Oh - I can just use the equations of motion here then, so I would have

x=x0+v0t+12at2​

x = x_0 + v_0 t + \frac{1}{2} a t^2

So here this is just x=0.5×100×9=450x = 0.5 \times 100 \times 9 = 450. So that I would have
moved 350350 meters during Δt\Delta t as a result of the force.

I think it may be simpler to use equations of motion. There's no reason these won't work with friction, as long as you include the friction force when you calculate acceleration.

What you know is:

1) A force (10N - friction) acts for 3 seconds. The block accelerates, reaches a speed $v$, and travels a distance $d_1$, say.

2) Friction slows the block from the speed $v$. The block travels a further distance $d_2$ until it stops.

You should be able to calculate the two distances from the equations of motion.

 

[gelfand]

So I have $a = 100, v = 300$ after three seconds.

Then I remove the force that's pushing the object and friction slows it to a stop.

Then at three seconds I have the kinetic energy

$$
\frac{1}{2} m v^2
$$

And this is all spent on work against friction (so that the object comes to a
stop ), giving

$$
F \Delta x = \frac{1}{2} m v^2
$$

So I have $F = m g \mu_k$ then

$$
\Delta x = \frac{\frac{1}{2} v^2 }{g \mu_k}
$$

Which is

$$
\Delta x = \frac{0.5 \times 300^2}{9.81 \times 0.2} \approx 22.935
$$

So that I travel around 22.8 km it seems! (really? )

Is this right? The approach etc.

Thanks

 

[PeroK]

So I have a=100,v=300a = 100, v = 300 after three seconds.

You're forgetting that the friction force is acting during the acceleration.

Which is

Δx=0.5×30029.81×0.2≈22.935​

Is that $22.9m$?

 

[PeroK]

Which is

Δx=0.5×30029.81×0.2≈22.935​

\Delta x = \frac{0.5 \times 300^2}{9.81 \times 0.2} \approx 22.935

So that I travel around 22.8 km it seems! (really? )

I did a quick calculation using $g = 10ms^{-2}$ and got exactly $1.8 km$. That includes the distance accelerating and decelerating.

 

[gelfand]

Sorry - that answer is correct but Its in thousands. I'd be unlikely to trust
any boss who gave me a job anyway tbh :)I thought that I had accounted for the frictional force during that, damn.

OK

$$
\text{Force net} = 10 - mg\mu_k
$$

Then we have that

$$
F_n = ma
$$

So

$$
a = \frac{F_n}{m}
$$

From this we can use impulse as

$$
I = F \Delta t = (10 - mg \mu_k) \times \Delta t = \Delta p
$$

So rearranging to solve for $v_f$ , where $v_f$ is the velocity at $t = 3$
seconds gives

$$
\frac{
(10 - mg \mu_k) \times \Delta t
}{
m
} = v_f
$$

We can use this value of $v_f$ to find the kinetic energy at $t = 3$ seconds

$$
E = \frac{1}{2} m v_f^2
$$We know that all this energy is spent on work to bring the object to a stop.

Work $= F \times \Delta x$, where $F = mg \mu_k$ so we have

$$
\Delta x \times mg\mu_k = \frac{1}{2} m v_f^2
$$

Dividing by $mg \mu_k$ gives

$$
\Delta x = \frac{\frac{1}{2} m v_f^2}{mg \mu_k}
$$

So we have to plug values here, velocity at 3 seconds is

$$
v_f = \frac{10 - 0.1 \times 9.81 \times 3}{0.1} = 294.114
$$

We have that the kinetic energy is equal to the work done which is $F\Delta x$,
where here $F$ is just due to friction and is $F = mg \mu_k = 0.1 \times 9.81 \times 0.2 = 0.1962$

Then we can sub this into the expression for $\Delta x$ as

$$
\Delta x = \frac{\frac{1}{2} (0.1)(294.114)^2}{0.1962} \approx 22044.60882
$$

So this is $\approx 22$ km

does that make sense?

thanks

 

[PeroK]

Sorry - that answer is correct but Its in thousands. I'd be unlikely to trust
any boss who gave me a job anyway tbh :)I thought that I had accounted for the frictional force during that, damn.

OK

$$
\text{Force net} = 10 - mg\mu_k
$$

Then we have that

$$
F_n = ma
$$

So

$$
a = \frac{F_n}{m}
$$

From this we can use impulse as

$$
I = F \Delta t = (10 - mg \mu_k) \times \Delta t = \Delta p
$$

So rearranging to solve for $v_f$ , where $v_f$ is the velocity at $t = 3$
seconds gives

$$
\frac{
(10 - mg \mu_k) \times \Delta t
}{
m
} = v_f
$$

We can use this value of $v_f$ to find the kinetic energy at $t = 3$ seconds

$$
E = \frac{1}{2} m v_f^2
$$We know that all this energy is spent on work to bring the object to a stop.

Work $= F \times \Delta x$, where $F = mg \mu_k$ so we have

$$
\Delta x \times mg\mu_k = \frac{1}{2} m v_f^2
$$

Dividing by $mg \mu_k$ gives

$$
\Delta x = \frac{\frac{1}{2} m v_f^2}{mg \mu_k}
$$

So we have to plug values here, velocity at 3 seconds is

$$
v_f = \frac{10 - 0.1 \times 9.81 \times 3}{0.1} = 294.114
$$

We have that the kinetic energy is equal to the work done which is $F\Delta x$,
where here $F$ is just due to friction and is $F = mg \mu_k = 0.1 \times 9.81 \times 0.2 = 0.1962$

Then we can sub this into the expression for $\Delta x$ as

$$
\Delta x = \frac{\frac{1}{2} (0.1)(294.114)^2}{0.1962} \approx 22044.60882
$$

So this is $\approx 22$ km

does that make sense?

thanks

My apologies, I had put a mass of $1kg$ in my "quick" calculation. I was trying to be too clever.

You're still forgetting the distance it travels during the acceleration.

 

[PeroK]

As an aside, let me show you a neat trick for these sorts of problems. You may or may not wish to remember this.

If an object accelerates from rest at a constant acceleration, then decelerates from this speed to rest at a constant deceleration. Then the average speed during both legs of the journey is the same: it's half the maximum speed.

You could solve this problem by working out the maximum speed, the time for each leg of the journey and the distance traveled would be half the maximum speed times the total journey time.

 

[gelfand]

As an aside, let me show you a neat trick for these sorts of problems. You may or may not wish to remember this.

If an object accelerates from rest at a constant acceleration, then decelerates from this speed to rest at a constant deceleration. Then the average speed during both legs of the journey is the same: it's half the maximum speed.

You could solve this problem by working out the maximum speed, the time for each leg of the journey and the distance traveled would be half the maximum speed times the total journey time.

Hrm ok - i'll think about how to apply that to this (i'm not sure on first read)

is the rest of the problem sound now though?

thanks

 

[PeroK]

Hrm ok - i'll think about how to apply that to this (i'm not sure on first read)

is the rest of the problem sound now though?

thanks

The working is correct, but you definitely need units. You'll lose marks for that in an exam.

 

[gelfand]

The working is correct, but you definitely need units. You'll lose marks for that in an exam.

OK cool - thanks for the help