# Tag Info

37

We tend to think that our modern electronic devices are very energy-efficient so mechanical mainsprings etc. must be enough but they're not. After all, the (Intel i7) microprocessors have over 1 billion transistors per chip and each transistor has to consume some nonzero (and not "totally" negligible) energy, after all, to do an operation and they do ...

31

Think on the equilibrium position: in figure A you have a force $mg$ exerted at the lower end and an identical force pushing the opposite direction exerted by the wall (if this force didn't exist, the spring with mass attached would just fall down by gravity). In figure B you have a force $mg/2$ exerted on the right end of the spring and an identical force ...

30

Start with figure A and ask what is the force exerted on the mass $m$ by the end of the spring? We know the mass $m$ is stationary, so the net force on it is zero, and we know there is a gravitational force downwards of $mg$. So the force exerted on the mass by the spring is an upwards force of $mg$ and therefore the tension in the spring must be $mg$. Now ...

25

Well, the sentence It seems like if it's an inherent property of the spring it shouldn't change, so if it does, why? clearly isn't a valid argument to calculate the $k$ of the smaller springs. They're different springs than their large parent so they may have different values of an "inherent property": if a pizza is divided to 4 smaller pieces, the ...

16

Assuming for a moment an infinitely hard and smooth surface, let's look at the energy of the ball. When the ball is dropped from a height $h$, initial potential energy is $mgh$. You would expect it to accelerate to a velocity $v=\sqrt{2gh}$. However, during the fall, it will experience drag from the air. This will cause the dissipation of some of the energy ...

14

Electrical analogies of mechanical elements such as springs, masses, and dash pots provide the answer. The "deep" connection is simply that the differential equations have the same form. In electric circuit theory, the across variable is voltage while the through variable is current. The analogous quantities in mechanics are force and velocity. Note that ...

12

When the ball hits the floor, its center of mass needs to be decelerated. It is done by deforming the ball (and possibly the floor if it is soft enough compared to the ball). In vacuum, if the ball and floor were perfectly elastic, they would then recover their previous shape following exactly the reverse dynamics as the deformation ones: in doing so, the ...

11

What an awesome question! By the way, as far as I know, the original video is here for those interested. One key to understanding this is the following fact from classical mechanics that is a version of Newton's second law for systems of particles: The net external force acting on a system of particles equals the total mass $M$ of the system times the ...

10

@HDE's comment's experimental approach done. Answer between about 1 and 2 ounces = 1/4 to 1/2 Newtons (http://www.wolframalpha.com/input/?i=convert+2+ounces+to+newtons). Key travel 3mm, say, = 1.5 milliJoules = $3.59×10^{-7}$ dietary Calories (http://www.wolframalpha.com/input/?i=convert+0.5+newtons+times+3+millimeters+to+Calories). So, for every ...

9

You are asking two questions really 1) How is PE actually stored in a steel spring at the atomic level? The explanation for this lies in quantum mechanics 2) Could you explain in detail how/where potential energy is actually stored in a steel spring and why the material never surrenders to the bending forces taking a new shape? Replying to 1) ...

8

You could make an analogy between the pressure distribution of a sound wave and the mass density distribution of a realistic spring undergoing vibrations, but it wouldn't give you the explanation you're looking for. As a matter of fact, that would be more like explaining a sound wave in terms of springs, rather than what you're trying to do, i.e. explaining ...

8

Yes, if you burn it. Neglecting the metal, a $25\ g$ wooden mousetrap at $15\ MJ/kg$ should yield about $375\ kJ$.

8

The potential energy only being defined up to a constant does not imply that potential energy differences only depend on differences in position. To see this mathematically, assume that a function $U$ has the property that $U(x_2)-U(x_1) = f(x_2-x_1)$ for some function $f$. Then if we take $x_2 = x+\Delta x$ and $x_1 = x$, and divide both sides by ...

8

Let me guess: you take the spring as it is and hang your objects, right? Then measure the displacement. Try to do the following: hang any arbitrary object so that the string will stretch a bit from its initial state. Then add you 100g and 200g objects to the initial mass and measure the difference in spring's length. I will be surprised if you won't get ...

7

Yes. Some of the elastic energy stored in the spring does work by moving lattice dislocations through the metal - this is the physical mechanism responsible for the plastic deformation of the metal spring - and is the reason the spring may be permanently deformed when unloaded, even when the grip position applied to the spring has remained fixed. Plastic ...

7

Let $E$ denote a quantity that does not change over time (from the first principle). Consider a ball with mass $m$ dropped from a height $h$. As the ball drops, its speed changes due to the gravitational acceleration $g$, reaching a final value $v$ at impact. Thus, we can infer that the quantity $E$ depends on these 4 parameters: $$E(m,H,g,V)$$ where $H$ ...

6

Kind of. The negative sign indicates the direction of the force exerted by the spring on the mass. If you pull the mass to the right, the force from the spring is to the left. Since they go opposite directions, there is a minus sign. The problem states an external force exerted on the mass displaces it, presumably to a new equilibrium. The spring ...

6

To supplement the answer by Luboš Motl, I will come to this problem from a Material Science point of view. What you mean by the inherent property of the string is not the spring constant, in fact, it is Young's modulus $E$, which only depends on the properties of a material of a body but not it's shape. $$E = \frac{\text{tensile stress}}{\text{tensile ... 6 The main problem is to determine what corresponds to zero mass of the harmonic oscillator. Remember that a fraction of the spring mass also participates in the motion. By introducing an intercept \beta, your friend takes into account that the true zero of the mass parameter m may be shifted from what you think it is. So an affine model T^2=Cm+\beta is ... 6 You seem to be more interested in energy than force. It is not possible to say "the force exerted on the keys thus far is enough to push a car five miles, or is equivalent to 100 kg of TNT" because pushing a car over a certain distance and exploding a certain amount of TNT aren't examples of a set force. The TNT has a certain amount of energy. Pushing the ... 6 I haven't used a mousetrap for several decades, but as I recall the moving arm is about 5cm long, so the tip moves 0.05\pi or about 0.16m. To get 17J of work the force at the tip of the arm would need to be 100N. I'm fairly sure the force isn't anything like that great. I remember being able to pull the arm back with one finger. I would guess the force is ... 6 Hooke's Law is frequently used to model multi-dimensional materials because the stress tensor is simple (linear). The full expression can be found on Wikipedia. The simplification for 2D is straight forward (drop any terms with a 3 in the subscript). Note that whether deformation in one dimension affects the others is a property of the material and shows up ... 6 You're missing a somewhat subtle point in your analysis. The block on the left in your diagram, where the spring is at its equilibrium position, is moving, so it has kinetic energy (which you're currently ignoring). I'll leave it to you to sort out what the speed needs to be and check that CoE holds. It needs to be moving because, if it were not, then there ... 6 Since the force is a function of distance, you need to integrate:$$F = kx\\ W = \int F\ dx\\ W = \int k\ x\ dx\\ W = \frac12kx^2$$Add signs as needed... Your work considered the force to be constant - and that's not how springs work. 6 There is tension in the spring. It it extended and hence there is tension! It is the centre of mass that falls with acceleration g rather then each individual mass. So the equation$$mg-T=mg$$is invalid. As the two masses fall they will oscillate (getting closer and further away) and the tension will cycle. Let us call the distance fallen by mass A, ... 6 In this equation \omega does not refer to the speed of angular motion, but the frequency of oscillation when measured in angular terms (usually radians/sec, but it can be degrees/sec). Frequency is usually measured in cycles per second (Hertz), but it is sometimes more conveniently measured in angular terms, when it is called angular frequency. The angle ... 5 Given a spring with spring constant k whose extension is in the direction x, the magnitude of the force that the spring exerts is given by$$ |F| = k|L-l| $$where L is its length and l is its equilibrium length. Now, imagine that the two masses are at positions x_1 and x_2 with x_2>x_1, then the length of the spring is given by L = x_2 ... 5 The second solution is there to allow for arbitrary start and stop times. Using standard trig identities you can convert an arbitrary linear combination of \sin and \cos into a time-displaced sinusoidal function:$$A\sin(\omega t)+B\cos(\omega t)=R\cos(\omega(t-t_0)), where $R=\sqrt{A^2+B^2}$ and $\tan(\omega t_0)=A/B$.

5

You may be imagining that if you push with constant force $F$, the spring will compress until the spring has such a resistive force. But since the spring was not counteracting that force, your constant force $F$ was accelerating the mass. Upon reaching the point where the spring has force $F$ as well, the mass does not stop but has a speed such that $KE = ... 5 The equilibrium position in this case is not where the spring is not stretched, it is actually stretched by a$\Delta x$amount with$F_{spring}(0) = k\Delta x$. So the spring force on point A is a little smaller than in point -A, since$ F_{spring}(A) = -k(A-\Delta x)$and$ F_{spring}(-A) = k(A+\Delta x)\$ so it compensates the "extra" force. You have ...

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