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... we have that the restoring force in the spring depends linearly on displacement. $$F(x) = -k(x) \\ k = \frac{-F(x)}{x}$$ This calculation of the force is wrong. Actually the force ($F$) has two parts: The restoring force of the spring ($-kx$) which is proportional to the current displacement ($x$). And the spring constant $k$ is still a constant. The ...


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To illustrate the independence of the value of the gravitational field one could set up the spring mass system on a horizontal table and with no friction present the period of oscillation would still be $2\pi \sqrt{\frac mk}$ provided the spring could undergo compression as well as extension. The reason for this independence is that the restoring force, $F$, ...


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Please note that mass $m$ and the spring constant $k$ are both the same on earth as they are on the moon. Consequently, the time period of the spring does not depend upon differences due to the acceleration due to gravity. Hence it will not change when it’s taken to the moon. I think the value of 𝑘 for a spring hung upside down depends on gravitational ...


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It would make it more difficult as the springs will provide atorque that would oppose the rotation of the tyre( that is why you have difficulty in driving with a flat tyre instead of a full pressurised tyre.) This is because the tyre tends to become flat hence demeaning the use of circular shape


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So in the non-driven damped harmonic oscillator: $$ \ddot x + 2\gamma \dot x + \omega_0^2 x = 0 $$ where the undamped resonant frequency is: $$ \omega_0 = \sqrt{\frac k m } $$ and $$ \gamma = \frac c {2\sqrt{km}}$$ is tha dampin ratio ($c$ is the scale between damping force and velocity). First of all, we see $\gamma$ decreases with increasing mass. This ...


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From the equation of motion for driven, damped harmonic oscillator results that the amplitude is givn by $$ A=\frac{F_m }{\sqrt{m^2(\omega^2-\omega_d^2)^2 +b^2\omega_d ^2 }} $$ where Fm is the aplitude of the driving force, $\omega_d$ is the driving frequency, $\omega$ is the oscillator's frequency and b is the damping factor. From this we can see that the ...


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As pointed out in the comments, you can take the average over time or position. With that said, over time, you would do (as suggested by user @JEB): $$\bar{F} = \dfrac{1}{t_2 - t_1}\int_{t_1}^{t_2} m\ddot{x}\;{\rm d}t.$$ If it's over distance, then you have (what user @Claudio Saspinski did): $$\bar{F} = \dfrac{1}{x_2 - x_1}\int_{x_1}^{x_2} kx\;{\rm d}x.$$ ...


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While the spring is going from $\Delta x$ to zero (the normal length at rest), the mass is being accelerated, and is in contact with the spring. Just after reaching the unstreched position, the spring starts to decelerate, and the mass detaches from it. We can use the definition of a conservative force as: $F = -\frac{dE}{dx}$, and calculate the average ...


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Worst case scenario is you can treat $x(t)$ as a distribution, then find the average value of the acceleration: $$ \langle \ddot x \rangle \equiv \frac{\int_{t=0}^{\tau}\ddot x(t)dt }{\int_{t=0}^{\tau}dt} = \frac{\dot x(\tau)-\dot x(0)}{\tau - 0} $$ with the fact that Newton's law gives: $$\langle F \rangle =\langle m\ddot x \rangle = m\langle \ddot x \...


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why we use circular motion to explain spring force? It is because the shadow of a body in uniform circular motion experiences a force exactly similar to that experienced by mass on a spring. In fig, the mass $m$ is rotating at a constant angular velocity $\omega$ in a circle of radius $A$. Even though the particle is experiencing a centre seeking force, the ...


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Consider a mass $m$ in uniform circular motion with radius $r$ and angular frequency $\omega$. It experiences a force of magnitude $F = mr \omega^2$ towards the center of the circle at all times. Its position as a function of time will be something like $(x, y) = (r \cos \omega t, r \sin \omega t)$ (use trigonometry and $\theta = \omega t$ to see this.) At ...


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This is an answer to the heavily edited question. Firstly calculate the velocity of the mass $B$ just prior to impact on $A$, from conservation: $$2mg\times 3x=\frac12 2m v_0^2$$ $$v_0=\sqrt{6xg}$$ The collision is inelastic but without external forces, so conservation of momentum applies. An 'instant' after the collision $A+B$ travel at $v_1$: $$2m\sqrt{...


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Or is it some other problem which caused this strange result which I got? The problem is that your “friction” force, $\eta m g$, always points in the negative $x$ direction. It does not behave like friction which always points in the direction opposite $v$. I'm mainly looking for an answer which discusses the break ups of the equation governing motion at ...


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Assume the velocity at impact of $M$ is $v_0$ (which can be calculated from the drop height but that's trivial). The collision is inelastic but without external forces, so conservation of momentum applies. An 'instant' after the collision $M+m$ travel at $v_1$: $$Mv_0=Mv_1+mv_1=(M+m)v_1$$ $$v_1=\frac{M}{M+m}v_0$$ Now calculate the kinetic energy $K$ of $M+m$:...


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Hint: Try drawing a free-body diagram. The horizontal force on the mass is spring force and charge(if the mass has charged). The direction of restoring force is in such a direction to restore the initial state of spring and the direction of coulomb force depends on the sign of charge.


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Your left forefinger is exerting a force on one end of the band. Your right forefinger is exerting a force on the other end. Each end of the band is exerting a force on the finger with which it is in contact. You can see this, because the ends of your forefingers have turned white and have temporary grooves in them! [Try not to suppose that the band is '...


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For a typical slinky (wiki) and comparing the gravity of earth and moon, this will be the case as outlined by the other answers. However, for a general spring (wiki) or weaker gravity fields, the spring may be in a state where it is fully contracted and the weight of its own mass is insufficient to stretch it at all. In that regime, gravity does not play a ...


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Yes it would be stretched out due to the incredibly powerful gravitational field and tidal forces. The energy comes from the gravitational field. The potential energy gradient of the gravitational field near the black hole event horizon becomes very high, meaning the end of the slinky will rapidly gain kinetic energy and stretch till it becomes "...


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Does the solution to: $$ \frac{dU}{dz} = \frac d {dz}[U_{spring}(z)+U_{gravity}(z)] =\frac d {dz}[\frac 1 2 k z^2 - \frac 1 2 mgz]=0$$ depend on $g$?


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If a slinky is hanging vertically in a gravitational field, the amount of stretch in any short section depends on the weight of the coil hanging below that section. Less gravity will produce less stretch.


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Simple answer yes, Think about taking two extreme cases : How much does a slinky extend in a gravity-free space? None at all How much would it extend if it was on perhaps Jupiter or even a black hole ?It should extend by a large amount. Gravity does play a role.


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The time required for the ball to reach the floor is determined by the acceleration of gravity. On the moon, gravity is weaker and the ball has more time to travel horizontally before hitting the floor.


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Supposing that something is not moving, then it won't move unless an unbalanced force is acting on it. Generally for a wall you can pull on it and there will be other forces (from the ground, the rest of a building, whatever) which will balance your pull, and it won't move. You could have a movable wall where you can pull it harder than the other forces - in ...


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I have named the forces 1 and 2 now suppose you are applying force 2 on the wall and wall 1 is fixed. First of all Spring force would not act until there is a extension in spring. So If you pull a wall attached to spring then the spring force would increase according to th law $$F=k\Delta x$$ where $k$ is spring constant and $\Delta x$ is extension in spring ...


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The spring force doesn't limit the degrees of freedom of a body connected to it. For instance if you have a mass on one end of the spring and the other is fixed then the body is free to do oscillations and if you given some horizonatal displacement then it's free to move in that direction too. There is no constraint on the body and so it's not a constraint ...


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If we assume Hooke's law holds, then for a spring constant $k$ with resting length $\ell$, the spring force is given by $F=\pm k(x_R-x_L-\ell)$, where $x_R$ and $x_L$ are the positions of the right and left ends of the spring respectively (the $\pm$ sign is to take care of which side of the spring you are looking at). Now, if you assume identical masses $m$ ...


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You started out really well with your expression for the spring force and stating Newton's second law. To add friction into the mix we need to realize that the damping is proportional to the velocity of the oscillating mass. We could liken it to a similar system where the spring itself is frictionless but it is oscillating in a viscous fluid. Writing out an ...


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I'm not sure how to resolve this equation whilst also taking friction into account. This equation is called the simple harmonic oscillator and it does not take friction into account. If you want to take friction into account then you need to use the damped harmonic oscillator. An explanation of the damped harmonic oscillator is here: https://phys.libretexts....


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Friction always acts to oppose the actual or impending motion. The maximum frictional force in this case is the coefficient of sliding friction times the normal force. However, this maximum frictional force occurs when the mass slips; for rolling without slipping, the frictional force is less and may not be constant. You can search this exchange for more ...


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The friction in this situation is static friction. It provides a torque which causes an angular acceleration (first positive and later negative), but it does not dissipate energy.


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Without actually solving it, I expect an oscillation during the constant force application. Remember that a constant force has only the effect of moving the equilibrium position (think of a vertical spring in a gravitational field). Thus, when you start applying the force, the spring will no longer be at equilibrium but "compressed" in an out of ...


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