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In detail, I came up with 424m for the stretched length of a spring in order to change the mass of an object by 10^-9kg which originally was 1 kg. Problem said, "is it feasible?"
In my opinion, there is no spring that can be stretched for this long, so it is not feasible. However, I'm not sure...
Here is the diagram:
I’ve only drawn the diagram and made equations for the sum of forces in the x-directions and y-directions:
##\sum F_{x}=(7 lbs)sin\beta+(8 lbs)sin\alpha=0##
##\sum F_{y}=(8 lbs)cos\alpha+(7 lbs)cos\beta-13 lbs=0##
This problem fascinated me in lower division physics. Find the 2-d equations of motion for a Slinky going down a flight of stairs (assuming the path of the slinky is planar; eg only going up and down and front and back, no side to side). I do confess that whilst I do love physics I’m not...
I know that from the given problem, I need to find the expression for Kinetic energy,
KE = 1/2 m [r(dot)]^2
and Potential energy,
PE = 1/2 k r^2
So L = 1/2 m [r(dot)]^2 - 1/2 k r^2
Hence H = 1/2 m [r(dot)]^2 + 1/2 k r^2
I assume that the fixed length r0 is provided to find the value of end...
Hi All,
I'm doing research in magnetic nanoparticles that are coated with chain molecules (oleic acid, I believe) and I am trying to model these molecules' effective spring constant.
The basic scenario is this: When a water-based ferrofluid is evaporated, it leaves behind only dried...
A mass (M) is attached to a spring (K). Mass moves in a one dimensional plane (horizontally)
1) If mass M is initially at x=0, what is the minimum Work required to bring it to x=x0 ? PE ?
2) M is released from x=x0, PE when x=xo/2 ? KE ?
3) PE when x=0 ? KE ?
4) PE when x=-x0/2 ? KE ?
5) What is...
I first find the force of friction to be (2)(9.8)cos(65)(.22), then I find the pull of gravity to be (2)(9.8)sin(65).
The full equation I set up to be: 0 = kx + force of friction minus the pull of gravity
This gives me the wrong answer, 0.44 . My free-body diagram is that kx and force of...
I suppose spring compression to be X when jumpping from 1m. Therefore gravational potential mgh=760(1+X) and my cal:760(1+X)=0.5kX²
why the solution manual state that 760=0.5kX²+760X
Hi guys,
I am working on design where I am using a torsion spring. And I need make an analysis and see the deflection of the walls. How can I calculate the forces that applied from Spring's legs to walls.
Thank you all in advance.
Hi folks,
I have an interesting problem here from the real world, it's a design i am working on.
So I have an object that is pressed by an hydraulic press with 50kN, let's call it F_before. Then I drive in a jig to fixate it. But the part that holds the jig has a limited stiffness. Hence if I...
I have attempted to draw a sketch of this but can't see how the data they gave me help to find time period
This is what value I have ended up getting but I believe is wrong
Much appreciated for any help!
I have a simple Fortran code for solving velocity and displacement of spring, the code works fine when writing all together without using a subroutine. but as I am learning Fortran I tried to do it using subroutine, but I keep getting errors
appreciate any help in advance.
program msd
implicit...
Hi Guys,
Forgive me, as it has been quite sometime since I have done my spring theory.
The problem I am having is the following:
This is the current situation:
- In a steel block I have a pocket depth of 15mm
- I have a compression spring with the following information:
Outside diameter is...
I found the equations of motion as
##m\frac{\mathrm{d}^2x_1 }{\mathrm{d} t^2} = -\frac{mg}{l}x_1 + k(x_2-x_1)##
and
##m\frac{\mathrm{d}^2x_2 }{\mathrm{d} t^2} = -\frac{mg}{l}x_2 + k(x_1-x_2)##
I think the k matrix might be
##\begin{bmatrix}
mg/l + k & -k \\
-k & mg/l + k
\end{bmatrix}##...
This is a common homework problem and I did find a post here that talks about it, but that post was closed to comments, so I am reproducing it to be able to ask a question.
We are, apparently, according to solutions I have found, supposed to recognize that it is an inelastic collision, since...
I honestly have no idea why I'm getting this question wrong, because it seems fairly straightforward.
I thought that treating the two blocks as one object would work, so 5 + 3 = 8. With Fg = Fs, ma = kx. Then, 8 x 9.8/1300 = x. x would be 0.06. That's been marked as an incorrect answer, though...
I found the amplitude of the simple harmonic motion that results (0.367, and I know this is correct because I entered it and it was marked as a correct answer), and assumed it would be the same value for the maximum compression since x(t) = Acos(wt). And, since the maximum value of cosine is 1...
Using conservation of energy,
0.5kx^2=mgh=mgx
0.5kx=mg
0.5kx=mg, x=0.15, m=9, g= 9.8
So isn't it k= 1176N/m?
For this problem, I understand that you can't use conservation of energy, but why? There is gravitational potential energy at the top and spring elastic energy at the bottom, and no...
Hello everyone,
First of all a very happy new year to everyone! And a big thank you to all the people who contribute to this forum, I have learned so much from here.
I am prototyping a design for a part that will be used in a consumer product. I am in the early stages of researching...
Hello, I am new on this forum, so if I make any mistakes please inform me. Thank you.
I wonder why I cannot use forces instead of energy conservation in this question.
The question is:
"How far (x = L) does the spring stretch before the masses stop moving? Express your answer in terms of m, k...
I'm working on a project where we have a mass (50 kg) sitting on a spring (350 N/mm) and are subjecting it to a sudden impulse (20g) along the spring axis to simulate a shock. We have the profile of the acceleration defined as:
##a(t) = x''(t) = P\cdot \sin^2 (\pi \cdot t / T)##
Where P (peak...
Good afternoon,
I have some doubts about the tension force suffered by a spring to which a mass is hung and which is making a simple vertical armoin movement. My doubt lies in the fact that at the bottom of the pier (where the mass hangs), the spring exerts the restoring force that is given by...
Good afternoon,I am preparing a laboratory report on the study of the oscillations of a spring and the following questions have arisen:The script asks us to represent the mass against the squared period, in this case, the slope will correspond to the spring constant divided by 4Π^2 and the...
Summary:: A block of mass m is dropped onto the top of a vertical spring whose force constant is k. If the block is released from a height h above the top of the spring,
a) what is the maximum energy of the block?
b) What is the maximum compression of the spring?
c) At what compression is the...
Hi. I have a application for a bistable spring mechanism. But my problem is that all of the ones i have seen require to go past the "midpoint" before it flips to the other side. Is there any such mechanism that flips over to position 2 before you get to the "midpoint" from position 1?
Example...
Summary:: Calculating the inclination angle
A stick is on two springs with spring constants D1=500N/m and D2=300N/m. Consider the stick is without mass and can rotate around the point E, which is distant from spring 1 with 0,1m and from spring 2 with 0,8m. A force F=100N pulls the stick up...
1/2 m v2 + mgh = 1/2 k x2
1/2 (5) (9) + (5) (9.81) x = 1/2 (200) x2
100 x2 - 49.05 x - 22.5 = 0
x = 0.779 m or x = - 0.289 m
Answer key says the answer is 0.289 m but in my opinion the answer should be 0.799 m because I take h = 0 at the position where the spring has maximum compression so...
Hi,
Could I please ask where I am going wrong with this very simple question:
Here's my answer (units implied):
A force of 20 extends the spring by 1/100 and so the Work Done in performing this extension is 20 * 1/100 = 1/ 5
Now, the work done in extending a spring is given by the formula...
T = 2π * √(2/300), T = .513 seconds.
If I divide it by 4/3, I get a final answer of .385 seconds of touch.
I know the box isn't attached for the entire oscillation, so T has to be divided. To me, it makes sense to divide it by 4/3 (when the box falls, the spring is compressed, hits...
Equations provided: for a spring pendulum and m replaced with L and k with g for the same pendulum, but with no weight attached.
Greetings
I tried solving this by stating that the length is 0,50m (since no length of the spring is given) and turning around the equation for the spring...
Why doesn't the incline angle play a role in changing the ##m## component of this equation?
##T = 2π\sqrt{\frac{m}{k}}##
FOR QUESTION 25, PART B:
ANSWER:
My Solution:
For the displacement graph, the gradient is crucial to predict the behaviour of the displacement of the block through time.
At 1: System is released - velocity is zero, considering forces acting on block, kx < mg, as block is observed to move downwards, and object is...
I'm all messed up on this problem. I see you can get the solution (74cm, as listed in the back of the book) by simply setting mgh=1/2kx^2, saying that h=x, and then adding 15 cm to that since that's the original length of the spring. This is the solved solution I was given. But now I think...
From what I understand, the force between two current-carrying wires can be calculated as:
$$\dfrac {F}{L}=\dfrac {\mu _{0}I_{1}I_{2}}{2\pi R} $$
Doesn't this mean:
$$\dfrac {F}{L}=\dfrac {\mu _{0}I_{1}I_{2}}{2\pi 1^{2}}=\dfrac {1}{2}kx^{2} $$ ?
$$\dfrac {\left( 4\pi \times 10^{-7}\right)...
Its a very basic problem and my friend suggested a solution that we should equate mg and kx ie mg=kx and just plug in m=8 and x=0.16 but i think that we should equate the energies like mgx=1/2kx^2 ie because at the point where mg will be equal to kx the mass will still have a velocity hence it...
Could I please ask for help regarding part (a) of this question:
If I get part (a) then part (b) will follow. So, here's my answer to part (a):
I'll be using the formula Elastic Potential Energy in a spring = (Yx^2)/(2a)
Where Y is the modulus of the spring, x the extension and a the natural...
a)
Our force can be represented as: $$\vec F= -k(r-H) \hat r$$ then the equations of motion are: $$\hat r: \ddot r -r {\dot{\theta}}^2=-\frac{k}{m_1}(r-H)$$ $$\hat{\theta}: r \ddot{\theta} + 2 \dot r \dot{\theta}=0$$
Plus we know that angular momentum is constant then $$|\vec L|=m r^2...
My solutions: When ball is launched horizontally, assuming its velocity is entirely in the horizontal dimension, there is no interaction of the ball with the gravitational field, thus no change in GPE, so all of the EPE (elastic potential energy ) of the spring is transferred to KE of the ball...
Hello All.
I am mentoring a high school student in my area with his class project for school. He has chosen he wants to launch an object (in our case, a softball) into a 5' diameter area. The idea is to build basically an oversized slingshot using an extension spring as the source of energy.
We...
Hello! I am stuck on part of a problem and was wondering what I am doing wrong. For part a of the problem, we were asked to find the impact speed. I did this in a photo below given the following values:
Θ = 30 degrees. The initial velocity = 10 m/s. The coefficient of kinetic friction = 0.4...
Hi everyone, just a quick question..
I tried this problem using Newtons laws, not conservation of energy, and I got an answer exactly half of what the correct answer is, and I'm not sure why. Here is what I did:
Net force = zero once the spring is compressed, therefore
mg - kx = 0
mg = kx...
Okay so, recently I got a job with my local newspaper delivering newspapers to make some money while deciding how I want to continue my educational career (I already have some college under my belt but I'm taking a semester off). All the newspapers have to be at the houses by 6 am so in order to...