Hot answers tagged

6

Let's make a simple example. A block with a compressed spring attached to it is on a frictionless horizontal surface against a stationary, immovable wall. The spring is released, and the block is then pushed away from the wall, thus gaining kinetic energy. The relevant forces here are 1) the force between the spring and the block and 2) the force between the ...


6

Biological systems are really tricky in physics class. Our intuition of how they work rarely lines up with how they actually do. In this case, the answer is almost correct. Cleonis is correct that there is some work due to pushing the earth, but it's negligable. We can really say that their answer is correct unless we're talking to a rules lawyer or are ...


5

Mathematically, moving between inertial and non-inertial frames correspond to moving terms from one side of Newton's second law to the other side. So, in your non-inertial frame accelerating with the incline you have for Newton's second law along the incline (using your notation) $$Mg\sin\theta+Ma_0\sin\theta=Ma_\text{net}$$ Moving to the inertial frame we ...


4

Warning: Non-rigorous math up ahead. Usually we handle instantaneous things in physics with the Dirac Delta function, which is qualitatively an infinite spike at some given point. We can use this here to express the instantaneous force in terms of an impulse per time: $$F(t)=J_0\delta(t)$$ so that the total change in momentum of the object is given by $$\...


4

When light is shined through a gas of Rydberg atoms the light experiences Rydberg atom mediated photon-photon interactions. That is, the photons are not interacting directly with each other. The cross section for any proposed direct photon-photon interactions (see the Schwinger limit) are orders of magnitude below what is experimentally accessible at the ...


3

The energy transfer is clearly internal but work must be done by the force as work done is defined as the (dot) product of force and displacement and the definition makes no reference to any transfer of energy. Yes, but you must be careful. What force and what displacement? In this case the force is the force from the wall on the skater’s hand. The ...


3

I think I agree there is something off in that quoted section. The purpose of that section is not practical application, the purpose is to bring an abstract concept into focus. And that means the usual simplifications for practical purposes should not be used. There is a physics joke that goes as follows: When Arnold Schwarzenegger does his push-ups, he is ...


3

I don't understand why reaction force would decrease at the start of a countermovement jump. The reaction force changes because of changes in the vertical acceleration of the center of gravity of the persons body throughout the counter movement, as follows. At position A, before the person body starts dropping, the net force on the person is zero, so that $...


2

Generally, "decay" refers to reactions where the initial state consists of only one object, so matter-antimatter annihilation is not decay. As an aside, whether or not a reaction "involves" all three types of "forces" (strong, weak, EM) depends on how small of a correction you're willing to look at. Generally, every interaction ...


2

This is because of greater shear strain you will experience if you run on the train while its accelerating. But us humans do not deform when subjected to this stress but they fall down. There are two factors involved here Shear force(assuming uniform acceleration) Without using complicated terms, let's assume that you're a passenger(A) watching another ...


2

It is sometimes called a definitional constant, but this term may not be universally understood. See for instance [1]. [1] T. Quinn, Physical quantities, in Metrology and Fundamental Constants, Proceedings of the International School of Physics "Enrico Fermi", Course CLXVI, 2007.


1

Velocity is the integral of acceleration with time, and if your acceleration curve has a finite value at $t=0$ and zero otherwise, the area under the curve is zero. This means, your first interpretation is correct, and the body will not move. The rule to remember is: $$ \Delta \text{(momentum)} = \text{(impulse)} $$ The change in momentum equals the impulse. ...


1

Having skimmed the Wikipedia page on the Magnus Effect, we can take it that the Magnus force, $F_m$, is proportional to the velocity of the cylinder, and perpendicular to it. Note that in reality it depends on other factors, like the cylinder's angular velocity and radius, as well as the density of the fluid (here air), but I will leave this for you to add ...


1

Electric potential is a scalar field: it has a value at every point in space, and its negative gradient is the electric field. Electrostatic (potential) energy is different: it is just a single value associated with a configuration of charges, in this sense it has no "gradient". However, there are ways you can make the energy into a scalar field. ...


1

First of all don't worry for the confusion that inevitably arises from the Newton's laws, it's normal! Maybe going forward you'll discover everything comes from variational principles and the puzzle will be a bit clearer. Momentum is not a force, it is a quantity that describes the state of motion of an object; in a classical situation it is the product of ...


1

I think you are essentially talking about impulse. You are right that $\int_0^t F(u)du$ would be zero if the force is applied for an instant, but any real force is applied for at least some time and so the integral cannot be zero. However, it may be very small as in your case where the force is not of a very large magnitude. The same goes with acceleration, ...


1

The fact that the equation doesn't tell you the acceleration is 0, doesn't mean it implies the opposite. If A doesn't imply B, that doesn't mean that not B is true. It might be that B is still true but you just can't infer it from A. In this case A is "the equation is satisfied" and B is "the acceleration is 0". The correct way of ...


1

Imagine the object is a cube of side $b$ (I am making this assumption, since you did not mention anything about the shape of the object and that simplifies the calculations) at a distance of $l$ away from the sink. Consider the pressure on two opposite sides of the cube. One facing the sink and the other away from it. That means that the pressure difference ...


1

Perhaps you can see it better with this figure. to apply NEWTON second law , you have to calculate the components of the position vector to the mass in inertial system. $$\vec{R}= \pm\begin{bmatrix} x_0 \\ y_0 \\ \end{bmatrix}=\pm \left[ \begin {array}{c} s\cos \left( \vartheta \right) \\ s\sin \left( \vartheta \right) + y \left( \tau \right) \...


1

To avoid confusion caused by gravity, we will assume the lab frame is in an inertial frame, floating in space far from Earth. $F = ma$ works in this frame. In these coordinates, there is no net force on an object that stays at $x = 0$. Working in the lab frame, you see the lift accelerated upward. Given no friction, the inclined plane exerts a normal forces ...


1

It was not specified the exact string supporting scheme, but I will risk and assume that OP had this in mind : Assuming mass-less string and according to Hooke's law and force equilibrium condition : $$ -k\, x_0 + m\,g = 0 $$ So the answer in this case is,- yes string restoration force is a normal force type which reacts to weight/pressure applied.


1

Will friction act in the south direction and the west direction respectively, or will friction act only in the south west direction combined and the block move in the north east direction with some acceleration ? As @Warrenmovic pointed out, the friction force will be in the opposite direction of the resultant force, i.e., $\sqrt 2$ N in the south west ...


1

This is how I deal with problem of finite friction. Assume friction is infinite and find the required friction force needed to resist all relative motion. Lets say this results into two co-planar friction forces $F_x$ and $F_y$. Calculate the magnitude of the required friction force $F = \sqrt{F_x^2+F_y^2}$ Compare the magnitude $F$ to the actual available ...


1

Friction always acts opposing the motion of the object. So if you have a resultant force in a particular direction, friction will always act in the opposite direction. So if your particle is moving north-west due to a force, friction will act to oppose that motion (hence south-east), hope that helps.


Only top voted, non community-wiki answers of a minimum length are eligible