Maximum distance a spring will stretch

In summary: You can consider the spring to be compressed 0.07 m when the putty is at rest on it. When the putty hits, the spring expands, and the frame starts to move downward. Conservation of energy can be used to solve for the downward motion of the frame.
  • #1
PirateFan308
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0

Homework Statement


A 0.150 kg frame, when suspended from a coil spring, stretches the spring 0.070m. A 0.200kg lump of putty is dropped from rest onto the frame from a height of 30.0cm

Find the maximum distance the frame moves downward from its initial position


Homework Equations


SEp+GEp+Ek = SEp+GEp+Ek

mgh+(0.5)(m)(v)2 + (1.5)(k)(x)2 = mgh+(0.5)(m)(v)2 + (1.5)(k)(x)2

initial momentum = final momentum


The Attempt at a Solution


To find velocity of the putty just before it hits the frame:
Ep = Ek
mgh = (0.5)mv2
(0.150)(9.81)(0.30) = (0.5)(0.150)(v)(v)
(v)(v) = 5.886
v = 2.426 m/s

To find the velocity of the putty and the frame after the putty hits:
initial momentum = final momentum
mv + mv = mv + mv
(0.200)(2.426) + 0 = (0.200+0.150)v
v = 1.386 m/s

To find k of the spring:
GEp = SEp
mgh = (0.5)kx2
(0.15)(9.81)(0.07) = (0.5)(k)(0.07)2
k = 42 N/m

To find the distance the frame moves downward:
mgh+(0.5)(m)(v)2 + (1.5)(k)(x)2 = mgh+(0.5)(m)(v)2 + (1.5)(k)(x)2
(0.35)(9.81)(h-0.07) + (0.5)(0.35)(1.386)2 + (0.5)(42)(0.070)2 = (0) + (0) + (1.5)(42)(h)2
(3.4335h - 0.240345) + (0.439) = 21h2
0 = 21h2 - 3.4335h - 0.199
h = 0.20887 or -0.04537
h - 0.07 = 0.1389 or -0.1153
I figure the height (because of my reference) must be negative, so it must be -0.1153m

Is this correct? I got a bit confused about h on the last equation ...
 
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  • #2
PirateFan308 said:

Homework Statement


A 0.150 kg frame, when suspended from a coil spring, stretches the spring 0.070m. A 0.200kg lump of putty is dropped from rest onto the frame from a height of 30.0cm

Find the maximum distance the frame moves downward from its initial position


Homework Equations


SEp+GEp+Ek = SEp+GEp+Ek

mgh+(0.5)(m)(v)2 + (1.5)(k)(x)2 = mgh+(0.5)(m)(v)2 + (1.5)(k)(x)2

initial momentum = final momentum


The Attempt at a Solution


To find velocity of the putty just before it hits the frame:
Ep = Ek
mgh = (0.5)mv2
(0.150)(9.81)(0.30) = (0.5)(0.150)(v)(v)
(v)(v) = 5.886
v = 2.426 m/s
You should have used m = 0.2, but since mass cancels, it makes no difference
To find the velocity of the putty and the frame after the putty hits:
initial momentum = final momentum
mv + mv = mv + mv
(0.200)(2.426) + 0 = (0.200+0.150)v
v = 1.386 m/s
this looks good.
To find k of the spring:
GEp = SEp
mgh = (0.5)kx2
(0.15)(9.81)(0.07) = (0.5)(k)(0.07)2
k = 42 N/m
no, you are off by a multiple. The frame is initially at rest in its equilibrium position at 0.07 m from the unstretched position of the spring.
To find the distance the frame moves downward:
mgh+(0.5)(m)(v)2 + (1.5)(k)(x)2 = mgh+(0.5)(m)(v)2 + (1.5)(k)(x)2
(0.35)(9.81)(h-0.07) + (0.5)(0.35)(1.386)2 + (0.5)(42)(0.070)2 = (0) + (0) + (1.5)(42)(h)2
(3.4335h - 0.240345) + (0.439) = 21h2
0 = 21h2 - 3.4335h - 0.199
h = 0.20887 or -0.04537
h - 0.07 = 0.1389 or -0.1153
I figure the height (because of my reference) must be negative, so it must be -0.1153m

Is this correct? I got a bit confused about h on the last equation ...
use the positive value of h, not the negative. Draw a sketch. Otherwise, your equation is correct after you change the k value.
 
Last edited:
  • #3
PhanthomJay said:
no, you are off by a multiple. The frame is initially at rest in its equilibrium position at 0.07 m from the unstretched position of the spring.

I am confused by this. Shouldn't I be able to use the equation

SEp+GEp+Ek = SEp+GEp+Ek

Where the left side of the equation is when the frame is raised and has no weight on the spring and the right side of the equation is when the frame is at rest 0.07 m from the un-stretched position of the frame. So Ek can be disregarded because the frame is at rest before and after so i would just use the equation

SEp+GEp = SEp+GEp

Since I am saying there is no weight on the spring before, and I am putting my height reference point at where the frame is at rest after, the equation is

GEp = GEp
mgh = (0.5)kx2

I am still missing where I messed up
 
  • #4
PirateFan308 said:
I am confused by this. Shouldn't I be able to use the equation

SEp+GEp+Ek = SEp+GEp+Ek

Where the left side of the equation is when the frame is raised and has no weight on the spring and the right side of the equation is when the frame is at rest 0.07 m from the un-stretched position of the frame. So Ek can be disregarded because the frame is at rest before and after so i would just use the equation

SEp+GEp = SEp+GEp

Since I am saying there is no weight on the spring before, and I am putting my height reference point at where the frame is at rest after, the equation is

GEp = GEp
mgh = (0.5)kx2

I am still missing where I messed up
The frame is at rest at 0.07 meters below the unstretched spring position because you slowly lowered it with your hand, applying a force until it comes to a slow stop. By applying a force, mechanical energy is NOT conserved, since you have done non conservative work. Only if you let the mass drop from the unstretched spring position would mgh =1/2 kx^2, and then x would not be 0.07 m.
 
  • #5
PhanthomJay said:
The frame is at rest at 0.07 meters below the unstretched spring position because you slowly lowered it with your hand, applying a force until it comes to a slow stop. By applying a force, mechanical energy is NOT conserved, since you have done non conservative work. Only if you let the mass drop from the unstretched spring position would mgh =1/2 kx^2, and then x would not be 0.07 m.

It doesn't say that the frame is slowly lowered by your hand. Couldn't you assume that it is simply dropped and that there are no external forces because it should end up 0.07m lower at rest whether you use your hand or not. If you didn't use your hand, i understand that it would drop lower that 0.07m but it would rebound upwards (so it would have kinetic energy and mgh would not equal 1/2kx^2), correct?

Also, I cannot think of another way to solve for k of the spring without using the conservation of energy. What other way is there in this situation?
 
  • #6
PirateFan308 said:
It doesn't say that the frame is slowly lowered by your hand.
It is implied. Otherwise, for an ideal spring, the frame would be bouncing up and down like a yo-yo, forever.
Couldn't you assume that it is simply dropped and that there are no external forces because it should end up 0.07m lower at rest whether you use your hand or not.
No, it will not be at rest at 0.07 m. In fact, it will have its max speed at that point, although the acceleration at that point will be 0.
If you didn't use your hand, i understand that it would drop lower that 0.07m
yes, to 0.14 m
but it would rebound upwards (so it would have kinetic energy and mgh would not equal 1/2kx^2), correct?[
It would rebound upwards, and mechanical energy would always be conserved (PE_s +PE_g + KE = constant at any point
Also, I cannot think of another way to solve for k of the spring without using the conservation of energy. What other way is there in this situation?
Hooke's Law!
 
  • #7
My second attempt at the solution:
To find velocity of the putty just before it hits the frame:
Ep = Ek
mgh = (0.5)mv2
(0.150)(9.81)(0.30) = (0.5)(0.150)(v)(v)
(v)(v) = 5.886
v = 2.426 m/s

To find the velocity of the putty and the frame after the putty hits:
initial momentum = final momentum
mv + mv = mv + mv
(0.200)(2.426) + 0 = (0.200+0.150)v
v = 1.386 m/s

To find k of the spring:
Fg = Fs
mg = -kx
(0.15)(-9.81) = -k(0.07)
k = 21.02 N/m

To find the distance the frame moves downward:
mgh+(0.5)(m)(v)2 + (0.5)(k)(x)2 = mgh+(0.5)(m)(v)2 + (0.5)(k)(x)2
(0.35)(9.81)(h-0.07) + (0.5)(0.35)(1.386)2 + (0.5)(21.02)(0.070)2 = (0) + (0) + (0.5)(42)(h)2
(3.4335h - 0.240345) + (0.336) + (0.051449) = 21h2
0 = 21h2 - 3.4335h - 0.1475
h = 0.1988m or -0.0353m
h - 0.07 = 0.1288m or -0.1053m

So the answer (which is positive) must be 0.1288m

I have put this as the answer, but it says it's still wrong. Any ideas where I went wrong this time?
 
  • #8
Ooops, I just realized that I should have added 0.07 m to the positive value of h, so it would be 0.1988 + 0.07 = 0.2688 m. It also says that this is wrong ...
 
  • #9
The 0.15kg frame has a weight of 1.47N and this causes an extension of 0.07m.
This gives a spring constant of 21N/m
A 0.2kg lump of putty dropped from 0.30m will have KE (= PE) of 0.589J
When the putty lands on the platform the spring will extend and this energy is stored in the spring.
The energy stored in a spring is 0.5F x ext or,
since k(spring constant) = F/e
F = k x ext then energy stored = 0.5 x k x e^2
The spring will stop extending when the 0.589J of KE have been stored (absorbed)
so 0.589 = 0.5 x 21 x e^2
This gives e (extension) of 0.237m
This is the maximum extension and the spring will oscillate.
 
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  • #10
EXTRA
when the oscillations stop the spring will essentially have a frame of mass 0.15kg with a 0.2kg lump of putty. The total mass is 0.35kg, weight = 0.35 x 9.81 = 3.43N and therefore an extension of 0.16m.
The falling lump of putty caused an extension of 0.237m over and above the original extension of 0.07m caused by the frame alone.
This means that the oscillations would be greater than the original extension and the whole thing would be bouncing around a bit.
 
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  • #11
PirateFan308 said:
My second attempt at the solution:
To find velocity of the putty just before it hits the frame:
Ep = Ek
mgh = (0.5)mv2
(0.150)(9.81)(0.30) = (0.5)(0.150)(v)(v)
(v)(v) = 5.886
v = 2.426 m/s

To find the velocity of the putty and the frame after the putty hits:
initial momentum = final momentum
mv + mv = mv + mv
(0.200)(2.426) + 0 = (0.200+0.150)v
v = 1.386 m/s

To find k of the spring:
Fg = Fs
mg = -kx
(0.15)(-9.81) = -k(0.07)
k = 21.02 N/m

To find the distance the frame moves downward:
mgh+(0.5)(m)(v)2 + (0.5)(k)(x)2 = mgh+(0.5)(m)(v)2 + (0.5)(k)(x)2
(0.35)(9.81)(h-0.07) + (0.5)(0.35)(1.386)2 + (0.5)(21.02)(0.070)2 = (0) + (0) + (0.5)(42)you forgot to change this value to 21.02!(h)2
(3.4335h - 0.240345) + (0.336) + (0.051449) = 21h2
0 = 21h2 - 3.4335h - 0.1475
h = 0.1988m or -0.0353m
h - 0.07 = 0.1288m or -0.1053m

So the answer (which is positive) must be 0.1288m

I have put this as the answer, but it says it's still wrong. Any ideas where I went wrong this time?
The problem is looking for the max stretch of the spring from the initial position of the frame. It's original position is 0.070 m beloe the unstretched length of the spring. Thus, you are looking for the positive value of (h - 0.07), once you correct the k value in your last term. Also, the problem has 3 significant figures in the given values, so round off your answer to 3 sig. figs.
 
  • #12
Wow, thank you so much PhanthomJay! You are a life saver!

So I got h=0.365 or -0.0384. Take the positive (0.365) and minus 0.07 and I get 0.295m. This is now correct?
 
  • #13
PirateFan308 said:
Wow, thank you so much PhanthomJay! You are a life saver!

So I got h=0.365 or -0.0384. Take the positive (0.365) and minus 0.07 and I get 0.295m. This is now correct?
It looks good to me! Are you entering your answer one last time ...3 strikes 'yer out?
 

1. What factors affect the maximum distance a spring will stretch?

The maximum distance a spring will stretch is affected by several factors, including the material of the spring, the diameter and length of the spring, and the amount of force applied to the spring. Additionally, the temperature and humidity of the environment can also affect the maximum stretch distance.

2. How can I calculate the maximum distance a spring will stretch?

The maximum distance a spring will stretch can be calculated using Hooke's Law, which states that the distance a spring stretches is directly proportional to the force applied to it. By measuring the force applied and the spring constant (a value that represents the stiffness of the spring), the maximum stretch distance can be determined using the formula d = F/k, where d is the maximum distance, F is the force applied, and k is the spring constant.

3. Can a spring stretch beyond its maximum distance?

Yes, a spring can stretch beyond its maximum distance if enough force is applied. However, this can cause permanent damage to the spring, altering its properties and potentially rendering it unusable.

4. How does the maximum distance a spring will stretch change over time?

The maximum distance a spring will stretch can change over time due to factors such as wear and tear, corrosion, and temperature changes. This can cause the spring to become less elastic and decrease its maximum stretch distance.

5. Why is it important to know the maximum distance a spring will stretch?

Knowing the maximum distance a spring will stretch is important for designing and using various mechanical systems. It ensures that the spring is being used within its safe limits and helps prevent accidents or damage to the spring and surrounding equipment.

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