# Tag Info

0

Surely the gravitational potential energy lost by the mass should equal the elastic potential energy gained by the spring? On the contrary, if something fell down doesn't that imply it gained some downwards velocity and hence the change in kinetic energy should be related to the net work done on the object? Here is a warning. Let's say you measured k ...

1

You have to make an additional assumption about how the friction force is affected by the viscosity. For small speeds you can assume Stokes flow around the oscillating mass. If the mass is a ball with radius $a$, then the friction force $F$ is given by velocity $v$ and dynamic viscosity $\mu$ by this relation: $$F = -6 \pi \mu a v$$ (see ...

0

Potential energy like Force occur in pair. If one has some potential energy due to 2nd, 2nd will have the same potential energy as the first. In the Gravitational potential energy equation : $U = \frac{GMm}{r}$ The potential energy is dependent on both the masses. This value is same regardless whether it is for 1st or 2nd. Both the body can do Same amount ...

3

Particles have gravitational potential energy due to its position in the gravitational field. Systems have potential energy. Ascribing the energy to a particle is incorrect. We say the particle has potential energy and not the Earth (the body doing the work). That is incorrect. The potential energy is a function of the system, specifically the ...

0

Potential energy is just energy stored in a static state -without motion. So a spring can have potential energy, and so can a body attached to the spring that's in a gravitational field. So for this type of system (undamped harmonic oscillator in a gravitational field) potential energy is not strictly defined for the spring. If the forces are conservative ...

1

The work-energy theorem is certainly the easiest way to do the problem, but you can also solve it by calculating the force. In any situation where you need to calculate the response of an object to a force you use Newton's second law. This tells us (after a minor rearrangement): $$\frac{d^2x}{dt^2} = \frac{F}{m} \tag{1}$$ In this case the force on the ...

1

Both are right. The first approach gives the compression where the net force on the object is zero. The second approach gives the compression when the velocity of the object is zero. When the block falls on the spring, it oscillates between $x=\frac{2mg}{k}$ and $x = 0$. Since the spring is ideal and the air resistance is negligible, this oscillation does ...

1

This is a simple system. You did not say about any further extention of the springs. And also the elevator moving with a constant acceleration. So in this case, the system is at equilibrium and weight act on the system is cancelled by the restoring force on spring. Now, as $m_1$ is in equilibrium, extention of second spring is due to massess $m_2$ and $m_3$ ...

0

Each side of the contact point has a 2×2 stiffness matrix defined in local (body) coordinates \begin{align} {\bf k}_1^{\rm body} & = \begin{vmatrix} k_1^{\rm long} & 0 \\ 0 & k_1^{\rm lat} \end{vmatrix} & {\bf k}_2^{\rm body} & = \begin{vmatrix} k_2^{\rm long} & 0 \\ 0 & k_2^{\rm lat} \end{vmatrix} \end{align} Now if each ...

2

If initially the mass is at $x=0$ and the initial velocity is $V$ then the (underdamped) position response is: $$x(t) = X \exp(-\beta t)\sin(\omega t) = \frac{V}{\omega} \mathrm{e}^{-\zeta \omega_n t} \sin(\omega t)$$ where \begin{aligned} \omega_n & = \sqrt{\frac{k}{m}} \\ \zeta & = \frac{d}{2 m \omega_n} = \frac{d}{2 \sqrt{k m}} \\ \omega ... 0 It seems to me you are making this more complicated than it needs to be. When the cable first becomes taut, the spring force is not yet in play and the only force will be v\cdot k - by the definition of the drag in the dash pot. You can compute the subsequent motion by solving the damped harmonic oscillator. Let me know if this is enough? 2 What you have here is two blocks: one of mass M and one of mass +\infty and a spring of mass m=0. So you have all the fun of zero mass and on top, now you have the fun of mass +\infty too. Newton's third law always holds. However, since the action-reaction pairs always apply to two different objects and only one of your objects has finite mass you ... 0 The force that the block applied to the spring will rebound back at the block, in the opposite direction, hence Newton's Third Law. Things don't necessarily happen at the same time, else the block wouldn't move (or always move with constant velocity) if we had two equal opposing forces simultaneously; there are energy conversions between potential and ... 0 Do not think in terms of force being applied by block on the spring. The K.E of the block is getting converted into the P.E of the spring and that is the only reason why the velocity of the block is decreasing. It is losing kinetic energy and finally comes to rest. If you think in terms of forces, then the force on the block by the tip of the spring is ... 4 Newton's third law applies. However the way it applies has been complicated by the fact that you described the spring as massless. If you had a spring with mass then when it exerts a force on something that something can exert a reaction force back on it. Since you gave the spring so little mass the force of the object on the spring just makes the spring ... 0 (v) I then differentiate this by v: t = d'(v) =\frac{mv}{k\sqrt{A^2-\frac{mv^2}{k}}} (am I crazy?) You made a mistake, \frac{d}{dv} d(v) doesn't equal time. For instance, \frac{d}{dv} d(v) can be the same at two different times. Consider the simplest case, a particle is at rest. Then v=0 and d(t)=const  so the derivative either ... 1 The obvious solution is to add a "dashpot damper" to the mix so that the equation of motion of the surface isy''(t) + 2 \lambda ~ y'(t) + \omega^2 y(t) = -\alpha w_0.As usual, in this case the equilibrium comes to a height $y_0 = -\alpha w_0 / \omega^2,$ which reads on the scale as the weight $w_0.$ Substituting $y = y_0 + \eta$ gives a function ...

1

There is the instantaneous force between the mass and the scales, and then there's the reading of the scales. Factors influencing both of these depend on many unknown factors - so here are just some general thoughts. First, if you drop a mass $M$ from height $h$ onto scales with mass $m$, and the two will then move as one, the collision is considered ...

Top 50 recent answers are included