Kinetic Definition and 1000 Threads

I want to experiment regarding the shock absorbing properties of different insoles. I do not know what formulas to use or how to experiment in the most reliable, valid and accurate way. Can someone please direct me in the right direction regarding my problem for both experimentations and...

Consider this situation. A mass $$m$$ is dropped onto the spring system from rest , and it is given that the smaller spring gets compressed by a distance $$h$$ My main query in this question is regarding the use of Work Energy theorem, which states that the work done on a body by all the forces...

Can someone prove kinetic energy is inversely proportional to potential energy?...I cant find any ans on internet , and I am unable to trove this.

How to calculate the kinetic energy of an object propelled by a magnetic field. For example I have 2 magnets face to face in opposition. Each magnet is a N42 Neodymium cube dimensions 10, 10, 10 mm they are in free space with no interference. The maximum pull strength is 4.7 kg. If I were to...

All three questions concern general relativity. 1. Does the curvature of spacetime depend on the frame of reference? 2. Does kinetic energy exist? 3. If it exists, does it contribute to the curvature of spacetime?

Is it possible to solve this without knowing the radius of the cylinder? My initial thoughts were that the energy required to stop it would be the sum of its rotational and translation kinetic energy, but I'm not sure it can be calculated without knowing the radius.

Proof of kinetic energy work done equals a change in kinetic energy in a mechanical system. δW = F.ds W = ∫F.ds W = m∫a.ds W = m∫(dv/dt).ds W = m∫v.dv here if v and dv are in the same direction the change in kinetic energy will be the usual equation. what happens if both are in different...

Attached to the post is a file with an attempt at a solution. The braking system is said to be a center-pull clamping system, which when activated by a brake lever (which equates to 355kN of force applied), triggers a tension force in the cables then clamps together two brake pads with spikes...

I think the angular velocity keep increasing on the plane with friction and the translational velocity keep decreasing due to friction while the total kinetic energy is conserved. When it moves to the frictionless plane, all energy converts to translational kinetic energy and it stop rolling...

Hello, I am having some confusions in what should be basic pointwise Newtonian mechanics, and would like to get some help with that. It is all about changing coordinates in potential energies. Let us start by considering a point particule in a 2d world with an axis x (left-right) and an axis z...

Can someone help me by taking a look at the attachment and figuring out where I am making a bad assumption? It's driving me nuts.

We know that net work done is equal to the change in kinetic energy, so we write: $$\Delta K = W$$ The tension is acting at angle ##\theta## due to the x axis, so we will only be taking its x component ##T_x = T \cos{\theta}##. Since we can look at this as one dimensional motion (##T_y## does no...

I was recently very surprised when I had a looked up relativistic kinetic energy. All sources gave the kinetic energy as the difference between total energy and rest energy, in some or other variant of the formula ##E_k=(\gamma−1)mc^2##. I didn't really understand at first. It seemed overly...

A few hours ago, I tried solving this problem and I'm still not quite sure if I've made a mistake somewhere or perhaps the guy in the video is wrong? Anyway, here's my solution: In the problem we're given that the total kinetic energy of of these 2 charges at this instant(look at the picture)...

I'm confused on this problem, as I feel they state two completely contradictory things in the explanation of how to solve it. The first statement that I feel contradicts the second is this: "We can see that the bullet’s speed v must determine the rise height h. However, we cannot use the...

Source: Shankar Yale OCW physics I have three questions here: 1. K_avg is 3/2kT, sure. But isn't this the kinetic energy of one particle only? So why isn't the answer multiplied by avogadro's number (because one mole). 2. When doing the "typical velocity" derivation, I noticed that they used...

Kinetic energy = 1/2 m V^2 I was thinking about this and thought another formula… E = mc^2 These look very similar except for the multiplication by 1/2. Let’s say you take a kilogram ball of uranium and accelerate it to the speed of light. I know, I know. You can’t. But let’s say you did...

FOr this, Use alternate coordinate system With ##ȳ##-axis parallel to incline and ##x̄##-axis parallel to the x-axis. Kinetic energy using this alternate coordinate system is ##T = \frac{1}{2}M\dot x_p^2 + \frac{1}{2}mR^2\dot \phi^2 + \frac{1}{2}m(\dot x̄^2 + \dot ȳ^2 + 2\dot x̄ \dot ȳ...

I expanded ET1=ET2 to get (Total energy at top) 1/2mv^2+mgh = 1/2kx^2 (Total energy at bottom) Rearanged i got k = (mv^2+2mgh)/x^2 so [(73)(20)^2+2(73)(9.8)(52)]/0.465^2 =479137.945N/m

I did some algebra and got the final equation mgh=1/2MR^2omega^2+1/2(2/5MR^2)omega^2, then plugged in the numbers and got an omega value of 111.09. Using 1.7 as h instead of cos(17)*1.7=1.63 I got 114.09. Both answers were wrong. Am I missing something, or did I just screw up the math?

part i) i did 1/2 * 1700 * v^2 i dont know what v is... so how do i solve it? part ii) i calculated it correctly by 440*25 please explain in detail why i used 440? and part d) i did 1.7*10^4 = 48000/t my t= 2.82 s but correct answer is 3.5s

Does the high temperature increase the kinetic energy of molecules or atoms, or does the high kinetic energy of atoms or molecules increase the temperature ? I'm so curious about this. Which concept is more accurate between the two

Homework Statement: Real world application of freshman physics Relevant Equations: TBD This is not a homework question, this is relevant to my work. It seems simple enough (introductory) but I keep running into problems. An electron is emitted from an surface (material is irrelevant, could...

I calculated the net force. I got 6500. I determined this is bigger than the force of static friction so the force of friction acting on the object must be kinetic. From there I got kinda lost. I know for the component to not slip Fnetx = 0 and Fnety = 0. But I'm not sure what to do from there...

Hello everyone, I'm currently working on a physics problem involving the rotation of a 5 kilogram ##M=5## solid sphere subjected to a force of 5 newtons ##F=5##, and I've encountered an inconsistency in my calculations. I'm seeking guidance or insights into where I might have gone wrong. My...

Hello everyone, I am working on this problem and I think I almost solved it, but then I noticed, that I do not know what values I have for dn, n and dθ. Can anyone help me with this?

It is good for the people who wants to know physics and it's beauty

Suppose in a different star system, a space shuttle sized spacecraft acquired a solar mass energy equivalent amount of kinetic energy, then passed through our solar system. While it was passing through the solar system would the craft’s gravitational effects be more similar to the space shuttle...

I tried E =Fxcos0 but only ended up with 243kJ

Let's consider a reaction A (reactant) -> B(product) and activated complex is denoted by C. This graph ( potential energy vs reaction coordinate ) tells us that reactant need some amount of activation energy (Ea) to convert in product, which has low potential energy which is shown here in...

For this, From the work kinetic energy theorem, if we assume that the book and the earth is the system, and that the finial and inital speed of the system is zero, then is the work KE theorem there is no net work done on the system. However, clearly there is work done on the system is shown by...

Given that there is a cylinder rolling without slipping down an incline, the method I was taught to represent the KE of the cylinder was: ##KE_{total} = KE_{translational} + KE_{rotational}## ##KE_{total} = \frac {1} {2} mv_{cm}^2 + \frac1 2 I \omega^2## Where "cm" is the center of mass, and...

The cylinder in question would have a moment of inertia of ~1.67kg*m² and rotational KE of 2.058J. At the point of impact also, assuming the body hits the sphere at a 90deg angle after traversing 90deg of displacement, it should(?) exert a force of 1.31N - enough to give an acceleration of...

Hello, I've seen in a few books on solid state physics that one can deduce an expression for average K.E.: $$<\:K.E.>\:=E_c+3/2\:k_B\:T$$ from the following: $$<\:K.E.>\:=\:\frac{\int \:\left(E-E_c\right)g\left(E\right)f\left(E\right)dE}{\int \:g\left(E\right)f\left(E\right)dE}$$ I can't...

I'm watching a video about " What is a black body?". That video said when the light interacts with the surface of a body, the electron and proton start oscillating. The electrons gain more transferred energy from the light that became its kinetic energy, rather than the proton because its mass...

I could not find any derivations in the litterature, except for the expected value of the energy flux expression itself: $$\overline{\Phi_{effusion,\epsilon}} = \overline{\dot{N_{ef}}}\overline{\epsilon_{ef}}=\frac{3Nl}{2A}\sqrt{\frac{(k_BT)^3}{2\pi m}}$$ I've started off by calculating the...

I would guess that by multiplying the pressure exerted by the shockwave on the body, and then the resulting force - here ~69 Newtons - per the distance the shockwave passed through when traversing body A, I could get the work done but I’m not sure if it’s that easy and whether or not I should...

W_ext is the external work done on B and C, which is 12 J Delta K_tot is the internal work, which is the work done by A on B plus the work done by A on C Delta K_tot = 5 Solving for \Delta U, we find that the change in potential energy is 7 J This answer says otherwise...

Dear Forum, I am solving for the expectation value of the kinetic energy for the deuteron (Krane problem 4.3). I must be missing something since this has become far more complicated than I remember. The problem is as follows: ## <T> = \frac{\hbar^{2}}{2m} \int_{0}^{\infty}...

I am seeing conflicting definitions of degree of freedom in my textbook. If I look at the definition given as per screenshot below then it is the number of independent terms/variables/coordinates used to define the energy of a molecule. But, if I look at the statement of Equipartition of energy...

First I found work: W=(3.85x10^5)(2.45x10^8) W= 9.43x10^13 Then used that for difference of kinetic energy: 9.43x10^13 = (1/2) (4.55x10^4)v2^2 - (1/2)(4.55x10^4)(1.22x10^4)^2 9.43x10^13 = (22750)v2^2 - 3.386x10^12 9.43x10^13 + 3.386x10^12 = (22750)v2^2 9.77x10^13 = 22750v2^2 9.77x10^13/22750...

I'm a little confused because my teacher used Bill's 500J of work for the kinetic energy equation and I don't understand why. I used the net work, so 300J, to find the speed and I'm not sure why that's wrong. Wouldn't friction make the wagon move slower than if there was no friction? So why...

hello guys, I wanted to ask whether I can just consider/think about this as being rotation around a fixed axis in a plane representing it as if it was 'just' a rod. This is mainly so that for the kinetic energy in the second position is where if we think about it in just a plane. Is this...

Kindly help me solve this question. The only thing so far that I know in this question is that energy is conserved and the momentum of Alpha particle will equal momentum of Thorium.

Spacetime expands at an accelerated rate and the particles with movement associated to this expansion are coupled to the Hubble flow. In many papers that I've read, objects coupled to the Hubble flow are treated as if they have some velocity and kinetic energy associated with it.However, can...

Young & Freedman 13th ed, Exercise 7.81 Starting with the crate, here is its free-body diagram: In accordance with Newton's First Law: $$ \Sigma F_y = 0 = n+(-w_c \cos{\alpha}) $$ Thus ## n = w_c \cos{\alpha} ##. And according to Newton's Second Law: $$ \Sigma F_x = m_c a_x = w_c...

Since there is no friction : $$ m \ddot{x} = 0 $$ (no x motion).For the kinetic energy , I've tried: $$ K = 1/2 I_{cm} \dot{\alpha}^2 + 1/2 m v^2_cm = 1/2 I_{cm} \dot{\alpha}^2 + 1/2 m \dot{z}^2$$ . Giving me a weird expression , shouldn't the kinetic energy just be half the the moment...

I tried just using the formula for kinetic energy but that was apparently the wrong answer. The answer key says it's (1/6)mv². I don't understand how they got that answer. Could someone explain?