Tag Info

New answers tagged

0

The first thing I should point out is W$\cdot$x. x is the displacement. Direction is important. That's really not the most important thing for your answer though. The punch line, Work is done if there's a force (hence change in velocity) and it causes displacement So, if something's moving a constant velocity, it doesn't have any force acting on it ...


0

A sailboat is moving at a constant velocity. Is work being done by a net external force acting on the boat? This is a bit of a trick question it tricks you into thinking there is a net external force. The boat is moving at a constant velocity; that's a given. That means that the net external force on the boat must be zero. But if I use Work = Force ...


0

Good one. The net forward thrust afforded by power of the engine, wind, rowing etc., is constantly overcoming and exceeding the retarding force of resistance by water, air, etc. So, the movement of boat in water and air is constantly impeded or constrained by water and air resistance-tending to bring the boat to halt. The forward thrust is constantly ...


1

I am not satisfied with the way @Bernhard answered, since it just shows the maximum velocity, thus only answering partially the question. The air resistance can be written as : $$ R = \frac{1}{2}\,C_x\, \rho\, S\, v^2 $$ Note : The mass of the object is not in this equation. This is very important. Applying Newton's law to one of the object gives at any ...


2

The wind is certainly doing work, because it applies a force and the point where the force is applied is displaced. However it isn't doing any work on the boat, it's doing the work on the water. The key point is that the net force on the boat is zero. We know the net force on the boat is zero because the boat is moving at constant velocity - if the net ...


0

The distance is a function of the angle. I am pretty sure you won't be able to obtain the inverse function analytically, not taking into account air resistance, so you should solve the relevant equation numerically, using one of the well-known methods, such as dichotomy or the secant method.


0

A couple things, first you are not discussing air resistance correctly. The drag depends on the current velocity, which is a dynamical quantity, not just on the muzzle velocity. You need to use the current velocity at any step of the calculation. Second, in broad terms, you can think of the problem you face as one of root finding. You have some function ...


0

To keep things simple lets say the tug of war rope is stationary, both sides are pulling with equal force. A and B are both people, A pulls left with 100N, B pulls right with 100N. Now in your second example we replace B with C, your immovable weight. A pulls left with 100N. Newtons law says that the block must be pulling right with 100N, otherwise the ...


6

The period of an elliptical orbit is given by: $$ T = 2\pi\sqrt{\frac{a^3}{GM}} $$ where $a$ is the semi-major axis. For a circular orbit of radius $r$ we have $a = r$. The two orbits you show do not have the same semi-major axis, so they do not have the same period. However if the elliptical orbit had $a^3 = 4r^3$ then the period of the elliptical orbit ...


0

A very crude estimate using the impact depth method: The density of the human body is almost the same as that of water, so you would expect that you'll lose most of your velocity after penetrating a depth equal to the width of your body (measured in the direction orthogonal to the contact area so if you go in head first, it will be the length of your body). ...


11

Other answers & comments cover the difference in acceleration due to friction, which will be the largest effect, but don't forget that if you are in an atmosphere there will also be buoyancy to consider. The buoyancy provides an additional upward force on the balls that is equal to the weight of the displaced air. As it is the same force on each ball, ...


0

The problem is that 9.81 m/s^2 is the acceleration during the fall, not the impact. Let's suppose the impact lasts 0.1 s when a ball hits the ground at 10 m/s, assuming inelastic collision (meaning, the ball doesn't bounce back), average acceleration DURING impact will be 10/0.1 = 100 m/s^2 rather than 9.81 m/s^2 and that's the acceleration you should be ...


0

Counting just the deacceleration suffered, you can get an idea (the duration of the impact is related to the viscosity of the fluid). Using the speed you found, the impulse is: $$J=\Delta p=-(80\times31.3)=-2504\,N\cdot s$$ Estimating a duration for the impact with the water of for example $0.3\, s$, the average force suffered is: ...


19

Ball 1 will drop faster in air, but both balls will drop at the same speed in vacuum. In vacuum, there is only the gravitational force on each ball. That force is proportional to mass. The accelleration of a object due to a force is inversely proportional to its mass, so the mass cancels out. Each ball will accellerate the same, which is the ...


41

I am sorry to say, but your colleague is right. Of course, air friction acts in the same way. However, the friction is, in good approximation, proportional to the square of the velocity, $F=kv^2$. At terminal velocity, this force balances gravity, $$ m g = k v^2 $$ And thus $$ v=\sqrt{\frac{mg}{k}}$$ So, the terminal velocity of a ball 10 times as ...


2

Impact on water is a very complex topic. Your simple calculation just figures out the velocity of a free-falling body after a 50 m drop. That just tells you the initial relative velocity of body and water surface. It doesn't tell you much about the force at impact, or whether the person survived. There are two things that might kill on impact: high local ...


0

Skipping the (high-school level) math for a moment, and apply a layer of common sense gives us this: Objects attract each other, inertia depends on mass. Given the mass of the earth is large one can reasonably assume it to be stationary for human-scale purposes. If we are dropping a moon-sized object we may want to consider the earth's motion, but it won't ...


1

Newtonian mechanics should be accurate enough. There are general procedures that can help answer the question "is it safe to ignore this effect." See, for example, dmckee's comment. This answer is more of a question-specific approach. The force on the "falling" object by Earth is equal in magnitude to the force on the Earth by the falling object. This can ...


1

Say you have a weight tied to each side a a rope which is strung over a pulley with friction. Here's a really easy way to see why the tensions on each side of the rope can't be equal. Imagine a really stiff pulley - in other words, ${\bf F}_{friction}$ is high. If that's the case, it'll be possible to balance unequal loads on this pulley system - i.e. a ...


4

You seem to be saying that friction couldn't speed it up, because nothing else is moving that fast. Well, how fast is it moving? We can imagine the gyroscope axis parallel to the z axis, and the casing to be aligned such that the x axis goes through it. If the casing is tipped slightly, the gyroscope resists that turning and one side of the shaft has firm ...


3

Actually, in your rotating torus your are mimicking gravity, which is point outward. I.e. the outside of the torus acts as a floor. The centrifugal acceleration would be $g\approx\omega^2 R$. This $g$ plays the same role as the gravitational acceleration on liquid or gas pressures under normal gravity conditions, so you can say $\Delta p = \rho g h$, or in ...


0

You can analyze this by imagining a tiny tiny gap between the two masses. Physically we have exactly that, as the electrons at the surface of the first block are certainly not in contact with the electrons at the surface of the second block. Then we have a series of two collisions: the first is the initial impulse, the second is the collision between the two ...


0

If you assume that the fluid is incompressible, I'm relatively sure that it could be shown that 100% efficiency is theoretically possible. You could use a variety of mental models to do this. I would prefer to think about conventional hydraulics lifting some weight. If you match the weight to the pressure, then you could raise an external object some (m g h) ...


0

Turbines (impellers) have fins that work more like aircraft wings and sails than the buckets of a old water wheel. They have have very high lift/drag ratios which translate the high efficiency. It is friction and heat loss that kill efficiency not the work of the turbine (how much it spins).


0

I'm going to just use $32^∘$ below; it doesn't make a difference. Your equation isn't correct. You should have $F_x−f_k = 0$ or $Fcos32∘−f_k = 0$. The x-component of F is $F_x = F.cos32^∘$. Writing $F_x.cos32^∘$ doesn't make logical sense. Why? Shouldn't I be able to use $∑F=0$ in this problem to find the answer? Yes, you can.


3

There are three possible forms of motion for a small object in a $\propto \frac{1}{|r|}$ potential: hyperbolic motion parabolic motion elliptic motion The first and second possibilities are unbound, so the object comes from infinity and will go back to infinity afterwards. The third one is bound, which means that the object cannot escape the ...


2

Objects in orbit come pretty close. If you don't mind venting the cabin or taking a walk outside, even air drag can be very nearly eliminated. All you have left are very small forces due to being in a non-inertial reference frame, and drag from the very, very thin atmosphere. Neither would be noticed without some very precise equipment. To reduce these ...


1

Have you heard of superfluidity? It happens when you cool liquid helium below about 2 Kelvin. The helium then will flow freely and without any friction. If you induce a current vortex in liquid helium, it will remain flowing until the end of time (however, you cannot draw energy from it, because the liquid is frictionless) or until it warms up again. So, in ...


0

If your friend's energy+ yours= F1, then you would see your own energy expenditure halved, which we know cannot be the case. If your friend helps you push the object, then you are no longer applying the same force, or (lazy answer) the force is no longer localised and motivates the part of the object most subject to friction. So, once the object is first ...


0

The glass bob will reach the ground earlier as acceleration due to gravity is independent of a falling body's mass. Being an insulator, no induced current is developed in it due to Earth's magnetic field.


2

There is, effectively, only gravitation and friction acting on the pack of gum. However, the friction is not that strong (it is mostly independent of the velocity of the book, and dynamic friction is weaker than static friction) and it doesn't have that much time to act. Hence it doesn't affect the momentum of the gum noticeably. This is very related to the ...


3

I'm sure everyone has had that concern when we encountered the definition for the first time, in school. There is a valid reason why this definition is still persisted with, despite the deficiency that you hit on. The most popular (and simple) forces in physics (also the ones with which we begin learning physics) are conservative forces, implying that the ...


33

If you're pushing a 10-ton truck and it's not moving, you are not doing any work on the truck because the distance $ds=0$ and the nonzero force $F$ isn't enough for the product $F\cdot ds$ to be nonzero. Your muscles may get tired so you feel that you're "doing something" and "spending energy" but it's not the work done on truck. You're just burning the ...


2

The proof is most easiest if we use the vector notation. We have $$\vec \tau = \int {d\vec \tau } = \int {(\vec r \times dm\vec g)} = \left( {\int {\vec r dm} } \right) \times \vec g$$ where I have used the assumption that near the earth $\vec g$ is constant. Now according to the definition of center of mass we have, $${{\vec r}_{cm}} = \frac{1}{M}\int ...


-1

Its a violation of Newton third law..because Newton third law only valid when 1)same kind of force ie when there is force of same nature So electromagnetic and block forces are not of same nature so Newton third will not hold


6

Backspin! Those shots in which the cue ball "draws" backwards after hitting the target ball involve backspin. Without backspin, the cue ball cannot reverse direction. Consider what happens when the cue ball is not spinning at all when it hits the target ball. The cue ball will come to a dead stop if it hits the target ball straight on. Think of Newton's ...


1

The direction the ball will take depends on the angular momentum. The velocity with which the ball moves or bounces backwards but the chief determinant is the spinning effect of the incoming ball.


5

First of all, if the collision is elastic, the distribution of momentum in between the components is completely determined by momentum and energy conservation! This statement is most obvious in the center-of-mass frame where the total momentum is zero and the two objects are moving in opposite directions. The momentum conservation (the total momentum is ...


2

This is low-Reynolds number particle sedimentation. It turns out that the problem is strikingly difficult, despite the simplicity of the setup and even of the equations (Stokes plus dynamics of pointwise solid particles). Check the webpage of E Guazzelli who's been working a lot on this. However, I believe you can get a fair rendering with simply a ...


1

It would crash into the earth because the Earth's gravitational field is not uniform and, even if said ring were to be perfectly positioned, ignoring the effects of wind, strikes from cosmic debris (not a lot that low in the atmosphere), change in mass of the ring (e.g. corrosion), change in shape of the ring (due to e.g. gravitational forces, heat ...


1

The angular momentum of the ring would stay constant. So if the ring were built on the earth and then lifted up to 100 feet and fastened together, it would would initially rotate at the same rate at the Earth's surface. The prevailing winds would probably disturb its orbit and I suspect it would eventually crash into the Earth. I think it would take extra ...


0

I'm guessing that you understand clearly the effect of precession here. The reason why the wheel starts to fall down is that when we explain the change in angular momentum of the wheel, we say that the angular momentum vector only changes in direction- right? But the angular momentum vector of the wheel doesn't only point outwards; it also points upwards (it ...


0

From a mathematical standpoint, any collision in which no mass is lost is described by two equations: Conservation of energy: $ m_1v_1^2 + m_2v_2^2 = m_1v_1'^2 + m_2v_2'^2 + E $ Conservation of momentum: $ m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' $ You know the masses and initial velocities, so this reduces to a system of two equations in three unknowns ($ ...


1

To parameterize the degree of inelasticity you use the "coefficient of restitution" which is 1 for elastic processes and 0 for inelastic processes. This is described by $$ \text{coef. of restitution} = c_R = \frac{\text{final relative speed}}{\text{initial relative speed}} = \frac{v_2 - v_1}{u_1 - u_2} \,. \tag{*} $$ This also tells you how to compute the ...


1

Perfectly elastic and perfectly inelastic collisions are just limiting cases on a scale of how much kinetic energy is retained. As noted in @Nathan's answer, if you work in the center-of-mass frame, a perfectly inelastic collision results in 0% of the kinetic energy retained, while perfectly elastic collisions have 100% of kinetic energy retained. So, you ...


3

There is not such thing as a "partially elastic" collision. Classical collisions between particles can be separated into two categories: elastic and inelastic. Elastic collisions are defined as collisions in which no energy leaves the system (i.e. $E_i = E_f$). All other collisions are inelastic, as some energy is lost ($E_i > E_f$). A perfectly inelastic ...


0

Neither, however... If you are standing on such a board, there is a simple way to propel yourself, assuming you can change direction (i.e. steer) the board. You cannot propel yourself forward, but you can propel yourself sideways, by pushing the board to one side. This gives you some sideways velocity. Then (before you fall over) turn the board so it is ...


0

You can't actually propel yourself forwards or backwards in this way (unless you are taking advantage of significant friction in the bearings). Moving your body forward or backward would cause the platform to move in the opposite direction but only so much as to leave the person+platform system's center of mass unchanged. Another consideration. Where would ...


0

Your derivation of the additional force on the scale for the falling rope is wrong, both cases yield the same results. If I understand correctly you are comparing the effect of a rope falling on a scale, to the similar fall of a liquid, something like this schematically: The problem of your reasoning is related to a misconception of the stagnation ...


3

Many students confuse the term work in physics with the conventional term of work. Your body wastes energy when you push something, and when that something doesn't move... 100% is wasted in the biological efficiency. 1st step: forget the concept of how hard it would be for you to do it. How much work is a table doing by holding up a 1kg weight? zero. It ...



Top 50 recent answers are included