# Energy Problem

## Homework Statement

"A spring of negligible mass and spring constant 60 N/m is attached to a block of mass 5 kg, which rides on a horizontal surface with a frictional kinetic coefficient of 0.15. A string then goes over a perfect pulley ( no friction or moment of inertia) and attaches to a hanging block of mass 2 kg. Initially, the spring is at the unstretched and uncompressed position, and the blocks are released at rest."

Ok... they first want me to find the a) fricitional force acting on the 5 kg block as slides to the right

and b) then the maximum extension of the spring

## The Attempt at a Solution

Ok.. What I really want is for someone to agree with the logic that I will state below in trying to solve these problems and if not, please correct me.

a) Ok.... So the only way I think I can solve this is through Netwon's second law right?

So for Block attached to spring, Fy= Fn-m1g=0
and Fx=T-f=m1a? Right?... if we consider the axes as they regularly are?
For Hanging Block.... Fx=m2g-T=m2a... if we treat the x as positive pointing down...
.... but if this is solve in this way, then how do I find the acceleration?... I mean I can cancel the two tensions.... Fx=m2g-f=(m1+m2)a... but how do I get the acceleration...

Or is it that simple... that I just do Fn=m1g and mutiple the frictional kinetic coeffiecient?

Or the second part...

I know we have to use the conservation of energy....

So Ei=Ef

That means, initially there is no kinetic energy or potentinal energy at rest

so... Ei=0???

so... Ef= 1/2kx^2+1/2(m1+m2)v^2+m2g(-y)+f(-y)=0..Right????

I mean the spring has potential energy, then the two blocks are moving at the same velocity as a system, so they have kinetic energy, and the hanging block moves down, so it has potential energy, and the frictional force disippates energy in the form of thermal energy? Is this right... and if so.... how do we factor in the unknown factors of the velocity of the system, and the distance the two blocks move?

Thank you.

PhanthomJay
Homework Helper
Gold Member

## Homework Statement

"A spring of negligible mass and spring constant 60 N/m is attached to a block of mass 5 kg, which rides on a horizontal surface with a frictional kinetic coefficient of 0.15. A string then goes over a perfect pulley ( no friction or moment of inertia) and attaches to a hanging block of mass 2 kg. Initially, the spring is at the unstretched and uncompressed position, and the blocks are released at rest."

Ok... they first want me to find the a) fricitional force acting on the 5 kg block as slides to the right

and b) then the maximum extension of the spring

## The Attempt at a Solution

Ok.. What I really want is for someone to agree with the logic that I will state below in trying to solve these problems and if not, please correct me.

a) Ok.... So the only way I think I can solve this is through Netwon's second law right?

So for Block attached to spring, Fy= Fn-m1g=0
and Fx=T-f=m1a? Right?... if we consider the axes as they regularly are?
For Hanging Block.... Fx=m2g-T=m2a... if we treat the x as positive pointing down...
.... but if this is solve in this way, then how do I find the acceleration?... I mean I can cancel the two tensions.... Fx=m2g-f=(m1+m2)a... but how do I get the acceleration...

Or is it that simple... that I just do Fn=m1g and mutiple the frictional kinetic coeffiecient?

Or the second part...

I know we have to use the conservation of energy....

So Ei=Ef

That means, initially there is no kinetic energy or potentinal energy at rest

so... Ei=0???

so... Ef= 1/2kx^2+1/2(m1+m2)v^2+m2g(-y)+f(-y)=0..Right????

I mean the spring has potential energy, then the two blocks are moving at the same velocity as a system, so they have kinetic energy, and the hanging block moves down, so it has potential energy, and the frictional force disippates energy in the form of thermal energy? Is this right... and if so.... how do we factor in the unknown factors of the velocity of the system, and the distance the two blocks move?

Thank you.
Part 'a' just asks you to calculate the friction force, f, which you have noted is as simple as f = u_k(F_n). Then you can solve for T using the 2 equations you have correctly identified, and once you solve for T, you can get the spring extension using Hooke's law without having to use the conservation of energy approach.

Part 'a' just asks you to calculate the friction force, f, which you have noted is as simple as f = u_k(F_n). Then you can solve for T using the 2 equations you have correctly identified, and once you solve for T, you can get the spring extension using Hooke's law without having to use the conservation of energy approach.

Wait... so... How can you use Hooke's Law, once you have the tension, to find the maximum extension?

T=F

F=-kx?????

... But even if I wanted to do it the other, way, though harder, would previouis energy equations be correct? or wrong? Thanks.

Wait... so... How can you use Hooke's Law, once you have the tension, to find the maximum extension?

T=F

F=-kx?????

... But even if I wanted to do it the other, way, though harder, would previouis energy equations be correct? or wrong? Thanks.

Question: Are the two masses attached by a spring or by a string?

Question: Are the two masses attached by a spring or by a string?

The two masses are attached by a string, but one of the masses is held in place by a spring unstretched and uncompressed....

The two masses are attached by a string, but one of the masses is held in place by a spring unstretched and uncompressed....

Ah, OK. You said in your first post that:

So for Block attached to spring, Fy= Fn-m1g=0 and Fx=T-f=m1a

which is true before the blocks are released from rest. However, note that f is the force of static friction which you are not asked to find. You are asked to find the force of kinetic friction. This, as PhanthomJay pointed out, is just the product of the coefficient of kinectic friction and the normal force on the mass.

To start solving part b, first determine the net force in the horizontal direction of the mass attached to the spring as the mass is sliding. Once you have this, ask yourself how you can determine the maximum extension of the spring from it.

Ah, OK. You said in your first post that:

which is true before the blocks are released from rest. However, note that f is the force of static friction which you are not asked to find. You are asked to find the force of kinetic friction. This, as PhanthomJay pointed out, is just the product of the coefficient of kinectic friction and the normal force on the mass.

To start solving part b, first determine the net force in the horizontal direction of the mass attached to the spring as the mass is sliding. Once you have this, ask yourself how you can determine the maximum extension of the spring from it.

.. So you're saying that I should find the acceleration of the system, then use it to find the tension in both ropes, to find the maximum extension??

So.... T=-kx???

But what if I wanted to do it using the conservation of energy???

.. So you're saying that I should find the acceleration of the system, then use it to find the tension in both ropes, to find the maximum extension??

Both ropes? What ropes? I thought there was only a spring and a string. Anyways, read what I said again.

So.... T=-kx???

No. Draw yourself a free body diagram of the situation. Analyze the forces and you should quickly see that what you wrote above is wrong.

But what if I wanted to do it using the conservation of energy???
I'm not sure that would help, although I may be wrong.

Both ropes? What ropes? I thought there was only a spring and a string. Anyways, read what I said again.

No. Draw yourself a free body diagram of the situation. Analyze the forces and you should quickly see that what you wrote above is wrong.

I'm not sure that would help, although I may be wrong.

..... Ok, so I see that for block #1 held in place by the spring,

Fx= T-f(kinetic friction)=m1a
Fy= Fn-m1g=0

.. for block #2..... m2g-T=m2a... but I still don't understand how you can get the extension, without conservation of energy????

Fx= T-f(kinetic friction)=m1a

Aren't you forgetting something, viz. the force by the spring?

Both ropes? What ropes? I thought there was only a spring and a string. Anyways, read what I said again.

No. Draw yourself a free body diagram of the situation. Analyze the forces and you should quickly see that what you wrote above is wrong.

I'm not sure that would help, although I may be wrong.

Aren't you forgetting something, viz. the force by the spring?

Oh.. so is it.... Fsp+T-f=m1a (box#1)
mg-T=m2a (box #2)......... But where would I go from there, since the acceleration, Fsp, and T are unknown values? Thank you.

Oh.. so is it.... Fsp+T-f=m1a (box#1)
mg-T=m2a (box #2)......... But where would I go from there, since the acceleration, Fsp, and T are unknown values? Thank you.

Question: In what direction is Fsp acting? Do you know about Hooke's law?

Question: In what direction is Fsp acting? Do you know about Hooke's law?

Hooke's law, F=-kx, so the Force of the spring is acting opposite to the tension of the rope? Right.. missing something? So, assuming this is right,...

T-Fsp-f=m1a
mg-T=m2a

... still if we don't know Fspring, tension, and acceleration are????

Hooke's law, F=-kx, so the Force of the spring is acting opposite to the tension of the rope? Right.. missing something? So, assuming this is right,...

T-Fsp-f=m1a
mg-T=m2a

... still if we don't know Fspring, tension, and acceleration are????

Good, except that your second equation should be m2g-T=m2a. You know what the acceleration is, you just have not realized it yet. (Think about the system when the spring is in its new equilibrium position.)

Good, except that your second equation should be m2g-T=m2a. You know what the acceleration is, you just have not realized it yet. (Think about the system when the spring is in its new equilibrium position.)

So... Hooke's law, says F=-kx=ma=But, since mass is neglected,

does, a=-kx?????

So....

T-Fsp-f=m1a, so

T+kx-(u_k*F_n)=m1*-kx????

So... Hooke's law, says F=-kx=ma=But, since mass is neglected,
does, a=-kx?????
So....
T-Fsp-f=m1a, so
T+kx-(u_k*F_n)=m1*-kx????

No no no. I guess I'll have to be more explicit: Consider the system when the masses are at rest again. What is the acceleration of the masses? What does the value of x represent when the masses are at rest again? Be careful with the sign in Hooke's law.

No no no. I guess I'll have to be more explicit: Consider the system when the masses are at rest again. What is the acceleration of the masses? What does the value of x represent when the masses are at rest again? Be careful with the sign in Hooke's law.

x represents the distance block #1 travels horizontally and x represents the distance block #2 travels vertically down, so are we trying to apply kinematics?, ....

v^2/x=a??? I really can't think of anything else as of the moment??

x represents the distance block #1 travels horizontally and x represents the distance block #2 travels vertically down, so are we trying to apply kinematics?,

Yes, that is true. I was hoping you would say something about x as it pertains to Hooke's law.

Yes, that is true. I was hoping you would say something about x as it pertains to Hooke's law.

... x is how far the spring is stretched/extended... where can I go from here?

... x is how far the spring is stretched/extended... where can I go from here?

You have not answered my question about the acceleration of the masses when they are at rest.

You have not answered my question about the acceleration of the masses when they are at rest.

... well, the acceleration of the masses at rest is 0.... so the acceleration is 0 right???? Ok... I understand what you are getting at, at the maximum extension, the spring can no longer extend, so the block stays stationary? Oh., so ....

Fx=T-Fsp-f=m1a=0 (Box #1)

Fx= m2g-T=m2a=0, so T=m2g,
so..

Fx= m2g-Fsp-f=0

m2g-f=Fsp ............ (m2g-f)/k=x..... Ok... that's tricky in understanding!!!

m2g-f=Fsp ............ (m2g-f)/k=x..... Ok... that's tricky in understanding!!!

Very good Very good ... Wait, but so I use the equation of

m2g-T=0, ok... so

m2g=T,....

And apply it to equation #1...

m2g-kx-f=0 ......... (m2g+f)k=x.... but my answer is half the answer in the back of book? What am doing wrong? Am I forgetting conceptual something??????

PhanthomJay
Homework Helper
Gold Member
... Wait, but so I use the equation of

m2g-T=0, ok... so

m2g=T,....

And apply it to equation #1...

m2g-kx-f=0 ......... (m2g+f)k=x.... but my answer is half the answer in the back of book? What am doing wrong? Am I forgetting conceptual something??????
Mind baffling, isn't it? Why not try your original work-energy formula, noting that v, as it was in the beginning, is now in the end, v=0, and also I note you've got a minus sign on the work done by friction, -f(y), it should be +f(x) when it's on the right hand side of your equation.... 0 = 1/2kx^2 - (m_2)(g)x +fx, solve for x.

Mind baffling, isn't it? Why not try your original work-energy formula, noting that v, as it was in the beginning, is now in the end, v=0, and also I note you've got a minus sign on the work done by friction, -f(y), it should be +f(x) when it's on the right hand side of your equation.... 0 = 1/2kx^2 - (m_2)(g)x +fx, solve for x.

... I got help from "e)h...", but he said that is was going to be harder to do so in terms of energy, but if the spring starts a position of x=0, and it is extended maximally a distance d from x=0, then isn't the maximum extension
2d, considering the length to the left, being d, of x=0??

PhanthomJay
Homework Helper
Gold Member
... I got help from "e)h...", but he said that is was going to be harder to do so in terms of energy, but if the spring starts a position of x=0, and it is extended maximally a distance d from x=0, then isn't the maximum extension
2d, considering the length to the left, being d, of x=0??
You're almost on the right track. When you use the Newton-Hooke approach that you and "e)h..." have nicely worked out, you get a spring extension that is exactly one-half the extension you get when using the work energy formula that I noted above. Read the initial problem statement very carefully. It asks for the maximum extension of the spring when the hanging mass is released from rest. Which solution is correct? HINT: Look at a simpler case of a mass hanging from a massless frictionless uncompressed spring . When the mass is released from rest, what is it's maximum extension?

You're almost on the right track. When you use the Newton-Hooke approach that you and "e)h..." have nicely worked out, you get a spring extension that is exactly one-half the extension you get when using the work energy formula that I noted above. Read the initial problem statement very carefully. It asks for the maximum extension of the spring when the hanging mass is released from rest. Which solution is correct? HINT: Look at a simpler case of a mass hanging from a massless frictionless uncompressed spring . When the mass is released from rest, what is it's maximum extension?

.... Maximum extension is how far the mass will move down, before the spring comes to its maximum extension..

But still... Why would I get half the correct answer???

... Wait, but so I use the equation of
m2g-T=0, ok... so
m2g=T,....
And apply it to equation #1...
m2g-kx-f=0 ......... (m2g+f)k=x.... but my answer is half the answer in the back of book? What am doing wrong? Am I forgetting conceptual something??????

You solved for x incorrectly. m2g-kx-f=0 -> kx = m2g - f -> x = (mg2 - f)/k.

You solved for x incorrectly. m2g-kx-f=0 -> kx = m2g - f -> x = (mg2 - f)/k.

Yeah.. you're right... but that was just a typo in the previous post, even when I do (m2g-f)/k... I still get the half of the correct answer. When I do it applying energy and work, I get the correct answer.... But still... why can't I use Newton's laws to get the correct answer. It makes no sense? Why is it half the answer????

PhanthomJay
Homework Helper
Gold Member
Yeah.. you're right... but that was just a typo in the previous post, even when I do (m2g-f)/k... I still get the half of the correct answer. When I do it applying energy and work, I get the correct answer.... But still... why can't I use Newton's laws to get the correct answer. It makes no sense? Why is it half the answer????
Using Newton's laws the way you have done, assumes that the hanging mass is slowly lowered to its new at rest equilibrium position, with no acceleration. And you calculate the unknowns based on this at rest position.The work energy method, however, assumes that the mass is released suddenly, in which case it will accelerate to a max speed then decelerate to 0 speed at the maximum extension, which is twice the 'equilibrium' extension. These are 2 separate assumptions. Looking again at the simple case of a 2kg mass hanging from a spring of spring constant k =60N/m, and using g=10m/s/s, then for the case where the mass is slowly released, using Newton, F_s = mg = 20N, and since x=F_s/k, x = 1/3m. But for the case where the mass is quickly released from rest, then using energy methods,
1/2kx^2 = mgx, solve x = 2mg/k = 40/60 =2/3m, which is the maximum extension of the spring, and twice the equilibrium extension.

Using Newton's laws the way you have done, assumes that the hanging mass is slowly lowered to its new at rest equilibrium position, with no acceleration. And you calculate the unknowns based on this at rest position.The work energy method, however, assumes that the mass is released suddenly, in which case it will accelerate to a max speed then decelerate to 0 speed at the maximum extension, which is twice the 'equilibrium' extension. These are 2 separate assumptions. Looking again at the simple case of a 2kg mass hanging from a spring of spring constant k =60N/m, and using g=10m/s/s, then for the case where the mass is slowly released, using Newton, F_s = mg = 20N, and since x=F_s/k, x = 1/3m. But for the case where the mass is quickly released from rest, then using energy methods,
1/2kx^2 = mgx, solve x = 2mg/k = 40/60 =2/3m, which is the maximum extension of the spring, and twice the equilibrium extension.

... 2/3 is not the answer. But I get what you are trying to explain. I think I have to relate work & energy.

Emech=-Etherm (Energy Dissipated by friction)

Or
Wnet=Wfriction

Ef-Ei=-f*x
(Uf+Ufsp+Kf)-(U+Uisp+Ki)=-f*x
(Uf +Usp+0)-(0+0+0)
1/2kx^2+m2g(-x)+0=-f*x....

... x(1/2kx-m2g+f)=0....

x=0... @ equilibrium position...

or 1/2kx-m2g+f=0....
there is where the half is.....

And I get 0.409 m (correct), which is unlike the .204 m, i got from the Netwon's Laws...

I just want to thank "e)hn..." and you, "PhantomJay", for helping me out with this problem and sticking with me. Thanks.

PhanthomJay
Homework Helper
Gold Member
... 2/3 is not the answer. But I get what you are trying to explain. I think I have to relate work & energy.

Emech=-Etherm (Energy Dissipated by friction)

Or
Wnet=Wfriction

Ef-Ei=-f*x
(Uf+Ufsp+Kf)-(U+Uisp+Ki)=-f*x
(Uf +Usp+0)-(0+0+0)
1/2kx^2+m2g(-x)+0=-f*x....

... x(1/2kx-m2g+f)=0....

x=0... @ equilibrium position...

or 1/2kx-m2g+f=0....
there is where the half is.....

And I get 0.409 m (correct), which is unlike the .204 m, i got from the Netwon's Laws...

I just want to thank "e)hn..." and you, "PhantomJay", for helping me out with this problem and sticking with me. Thanks.
Yes, the 2/3 and 1/3 numbers I used were just for an example using a different problem. Your numbers are correct, nice work! This problem required a bit of thought for sure.

Using Newton's laws the way you have done, assumes that the hanging mass is slowly lowered to its new at rest equilibrium position, with no acceleration. And you calculate the unknowns based on this at rest position.The work energy method, however, assumes that the mass is released suddenly, in which case it will accelerate to a max speed then decelerate to 0 speed at the maximum extension, which is twice the 'equilibrium' extension. These are 2 separate assumptions.

That is entirely correct. To reiterate: the x you calculated using Newton's laws is the x at the springs new equilibrium position. The x calculated using energy methods is the maximum amplitude (extension) of the spring when the masses are released. To find this latter x using Newton's laws would require fiddling with differential equations I think.