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36

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.


8

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 ...


6

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.


5

Because $W = Fd$ only holds for a very special case. The general definition of work is given via $$W = \int_\gamma \vec F (\vec r) \cdot \text{d}\vec r$$ where $\gamma$ represents a trajectory in $\mathbb{R}^3$ and $\vec F (\vec r )$ represents a vector field. The case where $W=Fd$ holds is when $\vec F (\vec r)$ is constant over all space and the trajectory ...


4

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....


4

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 "...


3

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 itโ€™s taken to the moon. I think the value of ๐‘˜ for a spring hung upside down depends on gravitational ...


3

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 ...


2

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 '...


2

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 $g$?


2

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 ...


1

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 ...


1

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 ...


1

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 ...


1

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{...


1

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 $M+m$:...


1

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.


1

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 ...


1

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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