Dynamic friction and spring force

[KFC]

Homework Statement

A light spring having a force constant of 125 N/m is used to pull a 9.50 kg sled on a horizontal ice rink. The coefficient of kinetic friction between the sled and the ice is 0.200. The sled has an acceleration of 2.00 m/s^2. By how much does the spring stretch if it pulls on the sled at 30.0 degrees above the horizontal?2. The attempt at a solution
The following is how I solve the problem by myself.

First, the object is only moving along the horizontal level and there is friction. I assuming the total force which pulls on the sled at 30 degree is F and the friction is f. The force affecting the spring is F_k = F_x = F\cos(30) - f.

For friction, it should be
f = \mu (mg - F\sin(30)) = \mu (mg - F_x\tan(30))

and from Newton second law
F_x - f = ma

so
F_x = ma + f
gives

F_x = \frac{ma + \mu mg}{1+\mu\tan(30)}

that is, according to Hook's Law

\Delta x = F_x/k = \cdots

I wonder if my calculation is corrected? I think about it for several times, it seems make sense. But one of my friend come to me few minutes ago. He said "the sled is experiencing a total force (along the horizon), which is F_x - f = ma, so the spring will also experience the same (total) force. Hence, x = ma/k instead of F_x/k". This seems also make sense, so which one is correct? WHY?

Here I attach the figure of the problem (given by my TA instead of from the text)

http://img697.imageshack.us/img697/4535/figt.jpg

 

[dr_k]

Homework Statement

A light spring having a force constant of 125 N/m is used to pull a 9.50 kg sled on a horizontal ice rink. The coefficient of kinetic friction between the sled and the ice is 0.200. The sled has an acceleration of 2.00 m/s^2. By how much does the spring stretch if it pulls on the sled at 30.0 degrees above the horizontal?


2. The attempt at a solution
The following is how I solve the problem by myself.

First, the object is only moving along the horizontal level and there is friction. I assuming the total force which pulls on the sled at 30 degree is F and the friction is f. The force affecting the spring is F_k = F_x = F\cos(30) - f.

For friction, it should be
f = \mu (mg - F\sin(30)) = \mu (mg - F_x\tan(30))

and from Newton second law
F_x - f = ma

so
F_x = ma + f
gives

F_x = \frac{ma + \mu mg}{1+\mu\tan(30)}

that is, according to Hook's Law

\Delta x = F_x/k = \cdots

I wonder if my calculation is corrected? I think about it for several times, it seems make sense. But one of my friend come to me few minutes ago. He said "the sled is experiencing a total force (along the horizon), which is F_x - f = ma, so the spring will also experience the same (total) force. Hence, x = ma/k instead of F_x/k". This seems also make sense, so which one is correct? WHY?

This is almost correct. The extension of the spring is not related to the F_x that you wrote, it's related to the the magnitude of F_spring:

http://img97.imageshack.us/img97/807/sledu.jpg

 

[shnav34]

A horizontal spring with stiffness 0.7 N/m has a relaxed length of 18 cm (0.18 m). A mass of 17 grams (0.017 kg) is attached and you stretch the spring to a total length of 21 cm (0.21 m). The mass is then released from rest. What is the speed of the mass at the moment when the spring returns to its relaxed length of 18 cm (0.18 m)?

I've attempted this problem several times, to no avail.

These are the equations I've been using.

Maximum speed = 0.5(mv^2) = 0.5(k)(x^2)
where k = spring constant and x is maximum stretch

V at any point = sqrt((k/m)*(xmax^2 - x^2))
k is spring constant, m is mass, x is equilibrium position and xmax is it's stretched distance

 

[dr_k]

A horizontal spring with stiffness 0.7 N/m has a relaxed length of 18 cm (0.18 m). A mass of 17 grams (0.017 kg) is attached and you stretch the spring to a total length of 21 cm (0.21 m). The mass is then released from rest. What is the speed of the mass at the moment when the spring returns to its relaxed length of 18 cm (0.18 m)?

I've attempted this problem several times, to no avail.

These are the equations I've been using.

Maximum speed = 0.5(mv^2) = 0.5(k)(x^2)
where k = spring constant and x is maximum stretch

V at any point = sqrt((k/m)*(xmax^2 - x^2))
k is spring constant, m is mass, x is equilibrium position and xmax is it's stretched distance

Good Evening shnav,

This is a different problem, so you'll need to start a new thread.

 

[shnav34]

ok keep an eye open for it, thank you, I didn't know

 

[KFC]

This is almost correct. The extension of the spring is not related to the F_x that you wrote, it's related to the the magnitude of F_spring:

http://img97.imageshack.us/img97/807/sledu.jpg

[/URL]

Thank you dr_k. May I ask you one question? What physics it related to such that the extension of the spring is related to F_spring instead of just the x component? My point is there is no stretch along y direction, so why can't we conclude that the y component of F_spring doesn't contribute to the extension? Thanks.

 

[dr_k]

Thank you dr_k. May I ask you one question? What physics it related to such that the extension of the spring is related to F_spring instead of just the x component? My point is there is no stretch along y direction, so why can't we conclude that the y component of F_spring doesn't contribute to the extension? Thanks.

http://img96.imageshack.us/img96/2180/springs1.jpg

For example, in the above picture, the block experiences a force, externally exerted by the spring, along the dashed line, its direction depends upon whether the spring is compressed or extended.

Calculate the magnitude of \rm F_{spring}. This magnitude is proportional to \rm k |\Delta s| , where \rm |\Delta s| is the one-dimensional compression/expansion of the spring along the dashed line.

For your problem, when you do your FBD, and apply Newton's 2nd postulate, you will, as you've done, break F_{spring} up into components. Then isolate and find the magnitude of \rm F_{spring} , which is then proportional to \rm k |\Delta s| .

 

[KFC]

Thanks. Just because the force should be along the dash line, the force exerted by the spring in my case should be F_x. Take the last figure given by you as example, the spring is compressed or stretched by along the dash line, hence, the force from spring should also along the dash line. And in the problem, the spring is along the x direction (horizontally), so the force of spring should also along this direction, right?

Or I put in this way, Hook's law is given by
\vec{F} = -k\vec{s}

where \vec{s} is along the same direction as the spring force (on which the spring compressed or stretched). Hence, if the spring is along x direction, the spring force should be \vec{F}=F_x \vec{\hat{x}}, not F_{spring} as shown in the figure given by you, right?

http://img96.imageshack.us/img96/2180/springs1.jpg

For example, in the above picture, the block experiences a force, externally exerted by the spring, along the dashed line, its direction depends upon whether the spring is compressed or extended.

Calculate the magnitude of \rm F_{spring}. This magnitude is proportional to \rm k |\Delta s| , where \rm |\Delta s| is the one-dimensional compression/expansion of the spring along the dashed line.

For your problem, when you do your FBD, and apply Newton's 2nd postulate, you will, as you've done, break F_{spring} up into components. Then isolate and find the magnitude of \rm F_{spring} , which is then proportional to \rm k |\Delta s| .

 

[dr_k]

Reading the problem that you posted, I envision a picture that looks like this:

http://img132.imageshack.us/img132/8369/springs2l.jpg

Does your text show a different picture?

Perhaps we're talking about two different pictures.

 

[KFC]

Oh, thanks for reminding me. I think my book gives me a wrong picture, the one shown in the book is different from your. And I read the problem carefully again, seems your picture makes more sense. Sorry to misunderstand you.

Reading the problem that you posted, I envision a picture that looks like this:

http://img132.imageshack.us/img132/8369/springs2l.jpg

Does your text show a different picture?

Perhaps we're talking about two different pictures.

 

[dr_k]

Oh, thanks for reminding me. I think my book gives me a wrong picture, the one shown in the book is different from your. And I read the problem carefully again, seems your picture makes more sense. Sorry to misunderstand you.

Can you draw the picture in your text, and post it? My comments in this thread are w.r.t. the picture I just posted. If the picture in your book is different than mine, then we were solving different problems.

 

[KFC]

Yes. I just add the picture in the first post. Actually, this is a practice given by my TA instead of my the text. I don't know where does this problem is from. I guess my TA gave me a wrong picture. Thanks dr_k, anyway.

Can you draw the picture in your text, and post it? My comments in this thread are w.r.t. the picture I just posted. If the picture in your book is different than mine, then we were solving different problems.

 

[dr_k]

Yes. I just add the picture in the first post. Actually, this is a practice given by my TA instead of my the text. I don't know where does this problem is from. I guess my TA gave me a wrong picture. Thanks dr_k, anyway.

Wow. That is not the picture I envisioned when I read

A light spring having a force constant of 125 N/m is used to pull a 9.50 kg sled on a horizontal ice rink. The coefficient of kinetic friction between the sled and the ice is 0.200. The sled has an acceleration of 2.00 m/s^2 . By how much does the spring stretch if it pulls on the sled at 30.0 degrees above the horizontal?

This pictures shows x(t) satisfying a non-homogenous 2nd-order ode, but since you're given the value of \rm \ddot{x}, the displacement is merely (for a stretched spring)

\rm \frac{F(\cos\theta + \mu_k \sin\theta)-m (\mu_k g+a_x)}{k}

if you know the magnitude of this F.

This is where a black-board would have come in handy, so we could have agreed on what problem we were trying to solve. :)


After this much work, why not solve the problem for this picture:

http://img132.imageshack.us/img132/8369/springs2l.jpg

 

[KFC]

OK. Here is what I did based on my picture. First, I am trying to find the total force on x direction

F_{xtot} = F \cos\theta - f = ma_x

here f is the friction, which is given by

f = \mu_k (mg - F\sin\theta)

Hence,
F_{xtot} = F \cos\theta - \mu_k (mg - F\sin\theta) = ma_x

From Hook's law, the force on the spring satisfies

F\cos\theta = - k x

namely,
F \cos\theta = \mu_k (mg - F\sin\theta) + ma_x = - kx

Hence,

x = \frac{\mu_k F\sin\theta - m(\mu_k g + a_x)}{k}

that close to what you gave except that there is no F\cos\theta in the expression, am I doining anything wrong?

 

[dr_k]

OK. Here is what I did based on my picture. First, I am trying to find the total force on x direction

F_{xtot} = F \cos\theta - f = ma_x

here f is the friction, which is given by

f = \mu_k (mg - F\sin\theta)

Hence,
F_{xtot} = F \cos\theta - \mu_k (mg - F\sin\theta) = ma_x

From Hook's law, the force on the spring satisfies

F\cos\theta = - k x

namely,
F \cos\theta = \mu_k (mg - F\sin\theta) + ma_x = - kx

Hence,

x = \frac{\mu_k F\sin\theta - m(\mu_k g + a_x)}{k}

that close to what you gave except that there is no F\cos\theta in the expression, am I doining anything wrong?

For the "new picture", we don't have a description of the problem, so we'd have to make up our own. So I'll make a few assumptions...

Suppose the block is moving to the right with increasing speed, and the spring is stretched \rm \Delta x from its equilibrium position:

http://img109.imageshack.us/img109/2784/springs3.jpg

Newton's 2nd Postulate implies:

\rm F_{{net}_x} = F \cos\theta - k\Delta x -f_k = m a_x

\rm F_{{net}_y} = F \sin\theta + N - mg = 0

Solve for \rm \Delta x