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0

You should be careful, since you need to take account of the force that the rain exerts on the trolley, or the momentum of the rain. Your second approach does this rather nicely, (with the assumption that the rain falls vertically, and hence doesn't contribute to the initial momentum). In the first approach, you could redo it to add the force that the ...


0

Newton's 2nd law in its most general differential form (ignoring vectors) is $$F_\text{net}=\frac{dp}{dt}=\frac{d(mv)}{dt}. \tag{1}$$ You must use the product rule on that bad boy $mv$. In method 1, it looks like you went straight to $F=ma$ which got you into trouble. You can really only use this simplified form of $m$ is constant or $a$ don't change; ...


2

For calculations of a collision, you look at (kinetic) energy and momentum equations. Momentum is conserved; energy may be dissipated. There is a direct exchange of momentum in the form of $F\Delta t$ - the same impulse that slows one object down accelerates the other, by the same amount. But the force may result in a deformation of the ball such that ...


1

Coefficient of restitution and hardness are not the same thing. The COR basically tells us how much energy gets lost in the collision process. In the case of a soft/hard ball with identical v1/v2 the COR=1, i.e. there is no energy loss. However, the collision of a perfectly elastic ball with a perfect wall will take a different amount of time, depending on ...


0

Many thoughts later, I think the correct free body diagrams are those : So, the vector equations are : Block : $$ m_2\vec{a_2}=\vec{W}_{weight-of-block}+\vec{F}_{plan-acting-on-block}+\vec{F}_{friction-from-plan-to-block}+\vec{N}_{normal-from-plan-to-block} $$ Plane : $$ ...


2

Yes - it really is that simple. There is no "upthrust" from the sand falling out of the car. There is just less mass in the car, and thus less force of gravity on the car. The sand gains momentum because gravity continues to pull on it after it leaves the car - but that no longer affects the car.


1

For simple harmonic motion, the restoring force is proportional to the displacement. So if you can compute the force $F$ on the drum when you displace it by a small distance $d$, then your "spring constant" $k = \frac{F}{x}$. I am sure you can compute that force (from the weight of the displaced fluid - Archimedes' principle). But warning - before you can ...


2

This looks like a homework problem, so I'll just give you a hint, rather than give you the whole answer. The hint is that according to Archimedes' principle, the buoyant force on a body that is fully or partially submerged in a fluid is equal to the weight of the displaced fluid. So the net force on the barrel at a given time is the weight of the amount of ...


0

While walking the work done by friction is zero. But who does the work, actually? How someone is getting displaced? This situation also arises when someone climbs without slipping or is climbing a ladder. Work is the transfer of energy by a macroscopic force. There is no energy being transferred between your feet and the ground, so you don't do any work ...


7

The diagram is misleading. Look at this: At any moment in time, you have the following forces on the particle: Gravity Tension in the string When you are at the bottom of the path, the tension in the string is equal to the tension needed to counter gravity, PLUS the tension needed to keep the mass in its path (in other words, to keep the string ...


5

Please note that in the picture, there are two forces acting: 1) the weight, mg, which acts vertically downward, and does not change, and 2) the tension in the string, Z, which points from the mass to the point the string connects to the ceiling, provided the string remains taut. Z varies with time periodically. These two forces combine to give the ...


-1

Prologue: There are friction while you walk (try do it over a very slippery surface and tell me the result). Also you legs are designed to move over irregular surfaces and do other amazing things like climb, swin and swing. By using a bicycle you can move forward over a flat surface in a more efficient way but most other things ill be a more difficult to do. ...


0

The angular velocity of the car equals the tangential velocity of the rear wheels divided by the radius of turn. $$\omega = \frac{v}{\rho}$$ The radius of turn is found from the steering geometry, but in general you can simplify it by $$ \rho = \frac{ \ell }{ \tan \theta} $$ where $\ell$ is the wheelbase of the car and $\theta$ is the steering angle. ...


0

Consider a simpler case, the earth is only twice as massive as the ball, and suppose they are 3 meters apart. (And suppose they are tiny, so their radius does not matter.) They fall together and meet at their common center of mass, which is 1/3 of the way from earth to the ball. So, the work done on the earth is F * 1 meter, and the work done on the ball ...


1

Just to expand on vaaaaaal's answer, let's simplify this very slightly by assuming that the ball falls at it's average fall velocity $v$ for the whole height $h$ over a time $t$. Obviously, $v = \frac{h}{t}$. Then, we know that total momentum is conserved, so the Earth must fall up with speed $v_{e} = \frac{m}{M_{e}}v$. Thus, over the whole time of ...


2

You really, really have to be able to draw a diagram to understand many physics problems. Here is my interpretation of what you write above (with apologies for poor lining up of various lines due to the limitations of the drawing package I had to hand). The normal force acts at right angles to the surface and is labelled as $N$. The weight of the ...


1

The force from the earth on the ball and the force from the ball on the earth are in fact opposite and equal but the amount of work done on each is not the same. The earth is much more massive than the ball so, for an equivalent force, it is going to accelerate much more slowly and move a much shorter distance during the time the ball is falling than the ...


0

I can't click the comment button so I will post here: there is still friction but unlike tires it doesn't oppose but it holds your feet without which you might slip ay every step. The work is done by your muscles, but if you go deeper it is your muscular system controlled by your nervous system being supported by your skelet system and maintained by your ...


2

The first the expression $U(r) = -\frac {GM} r$ is a potential, but not potential energy. The units are velocity2. This is a widely used potential in solar system astronomy, geology, geophysics, and in aerospace engineering. For example, see ...


1

Even just walking on flat ground is doing some work in the physics sense. Your center of mass will bounce up and down with each step. The up part requires work to be done, and the body has no mechanism to derive energy from joints being moved by external forces, so can't recover the work on the way down. At best the body could be a spring, which happens ...


2

Notice that $h$ and $r$ are related in the following way: \begin{align} r = R + h \end{align} where $R$ is the radius of the Earth (the distance from the center to the surface) and $h$ is the height above the surface. Then notice that \begin{align} U = -\frac{GmM}{r} = -\frac{GMm}{R+h} = -\frac{GMm}{R}\left[1 -\frac{h}{R} + O(\frac{h}{R})^2\right]. ...


1

Your muscles do some work when you start to get you up to walking speed. According to physics there is no work being done once you get up to a constant speed. At a constant speed there is no acceleration, no force, and no work. Unfortunately, just because you aren't doing any work doesn't mean your muscles aren't consuming any energy, muscles are inefficient ...


0

The entire system will be balanced and not move, as the momentum is conserved. There however will be an small shake back and fourth. First when it fires so the momentum move the rocket and the next one where the gas will hit the other side and conserve momentum and move it back to original location of starting point.


2

Static friction force arises whenever there is interaction between two bodies by direct contact (touch). There need not be any mutual motion between the bodies. This friction force is necessary to explain why the bodies around us maintain their position so reliably. Without friction forces, there would be nothing opposing their mutual motion and the world ...


5

According to Newton's 3rd law of motion, the force exerted is equally distributed between the rope and the ground. Force on the ground can never be greater than force on the rope and vice versa. Therefore both players exert the same force on the rope and ground, that means that the same amount of work is done on the rope and on the ground. The winner ...


6

The force on rope is equal for both of them at any time. For winning the game the force on ground is responsible.


0

Are you wondering why your two equations $W=\frac{1}{2}kx^2$ and $W=Fx$ don't match? That's because the latter equation, $W=Fx$, is only true for a constant force. The more general expression is $W=\int \vec F \cdot d \vec x$, similar to what you wrote in differential form $dW=F\,dx$. The expression with $\frac{1}{2}$ in it is correct. The expression with ...


1

For a book lying on a table, for example, the weight is cancelled by the upwards reaction force from the table. That's not quite true -- unless the table in a vacuum chamber at the south pole. The upward normal force exerted by the table and the downward gravitational force exerted by the Earth don't quite cancel. The book rotates with the Earth, and ...


0

I'm surprised none of the answers mention the phrase "normal force" which is what, I believe, you are asking about. JohnRennie's answer covers the concept, but doesn't use this term. Essentially, the normal force is the force exerted that resists gravity - you are pulled towards the earth by the force of gravity, the earth pushes back on you (keeping you ...


0

To 'derive' conservation of Energy from $ \vec{F} = m \vec{a}$, we take a dot product ($\hat{i} \cdot \hat{i} = 1$) which means that we have one (scalar multi-variable non-linear differential) equation with potentially many unknowns. Energy is nice because it provides a common language with all the physical sciences, but in classical mechanics, it's mostly ...


3

Suppose the Earth wasn't rotating, but instead you impart a small sideways velocity to the pendulum bob as you release it. What you would have is a conical pendulum that traces out an ellipse instead of a straight line. Now start the Earth rotating. The point of the Foucault's pendulum is that the rotation of the Earth doesn't affect the motion of the ...


1

Nothing happens to the initial angular momentum. It is simply irrelevant. Let's imagine that the pendulum is placed on a non-rotating body (or at the earth's equator). When the string is cut, the pendulum falls straight to the center. The path it traces is a straight line. In the non-equatorial case, the pendulum structure has some rotation. At the ...


4

My knowledge is limited on the subject but matter is typically prevented from collapsing under the weight of extreme gravity by particle degeneracy. This is what keeps neutron stars from collapsing into black holes and is the result of particles resisting occupying the same quantum states. There are also some recent observations that indicate that there is ...


14

Suppose you're standing on a box as shown in (a) below: There are four forces acting. You apply a downward force $mg$ on the top of the box, and by Newton's third law the box applies an upwards force $-mg$ on you. The box transmits your force to the ground, so the box applies a downwards force $mg$ on the ground and the ground applies an upwards force ...


0

You do have to define the velocity with respect to something. The idea (I think) behind defining wrt fixed stars is that the Earth is in motion wrt the stars, too. Barring friction, the moment the rope is burned, the pendulum is only under the influence of gravity and the restraining force of the suspension rope -- analogous to swinging a weight around ...


21

Yes, every gravitational force in Newtonian mechanics has an equal and opposing force, and it usually acts on other mass. More specifically, every two pairs of masses feel a gravitational force that's proportional to the product of their masses and inversely proportional to the square of their relative distance, but more important is the fact that both ...


0

I've heard that inertial frames are frames in which Newton's laws hold. The modern view of Newton's first law is that it defines the concept of an inertial frame. It also, at least conceptually, provides a mechanism for testing whether a frame of reference is an inertial frame. Suppose you know that no forces act on some particle. If that particle ...


0

1)Definition: An inertial frame of reference is a frame of reference where Newton's first law applies (uniform motion if without external force). Now if we have other frame of references that are moving relative to this inertial frame with uniform relative velocities, then all the others are also called inertial frame of references. 2)Transformation between ...


1

Friction does not depend on velocity (unlike viscous drag). An object that is stationary on a table will continue to be stationary when you push it gently - because there is an opposing force of friction. So no, your understanding is wrong: friction is present even when the object is just starting to move. Let me draw a diagram: That ought to clear it ...


0

Physicist created momentum as the property of the system that is conserved if on the system act only internal forces. A force is defined internal in the system if, the force itself and its reaction are applied on the system. From the previous hypotesis you can demostrate that momentum is defined as $$\vec{p} = \sum m_i\vec{\upsilon}_i$$ The last example ...


0

An elastic collision means that the over all kinetic energy of the entire system before and after the collision is the same. So the ball can bounce off the wall, and the wall can recoil in such a manner that you have an elastic collision.


3

Momentum is conserved in magnitude and direction. So in order to analyze any situation of momentum conservation, you should always start with $$ \sum \mathbf p_{i}=\sum\mathbf p_f $$ where the subscripts denote the initial and final momenta. As to the ball & wall, you are correct that momentum is not conserved if you are only looking at the ball. If you ...


2

It is the momentum of the entire system that is conserved. The fundamental reason for this is that the laws of physics are the same everywhere in space. This argument for momentum conservation is called Noether's Theorem. So where did you go wrong in your original example? Well you assumed that the wall was completely rigid. In reality that isn't actually ...


-1

The question "Why does a slow-moving (or stationary) bicycle fall over?" is kind of a boring one. A stationary bicycle falls over because it is at an unstable equilibrium. Specifically, a rigid body standing on a surface is at an equilibrium if its center of gravity is above the convex hull of its support (the points where it contacts the surface). If ...


-2

I agree with Brandon's analysis. To add some information on the "why we fall" question: The bike is unstable in any case in the absence of a driver. To discuss stability with a driver, you must model the driver in some way. In controlling it is usually modeled by some linear function. There is a function for any speed > 0 that stabilizes the bike, but that ...


0

Okay, let's first review the initial setup: 1st particle: mass $m$ and initial velocity $v_m>0$ in +x direction 2nd particle: mass $Am$ with assumption $A>0$, and initial velocity $v_A=0$ 3rd particle: mass $Bm$ with assumption $B>0$, and initial velocity $v_B=0$ Notation: after each collision, the new velocities will have a prime added to ...


1

The force of gravity will simply move the equilibrium point, being a costant force. So if $l_0$ is the initial lenght of the spring, the applying the orizontal component of the force $Mg\sin \theta$, you'll get the new lenght in equilibrium that is (Hooke's Law) $$ \vec{F} = -k\vec{(x - x_0)} \Rightarrow x = F/k + x_0$$ $$x_1 = Mg\sin \theta / k + x_0$$ ...


14

Alright I'll throw my hat into the ring with an answer. The idea that it's an unsolved problem is totally bogus. When you start to fall to one side or another if you turn the wheel slightly in the direction you're falling the bicycle starts to follow a curved path. There is a force due to friction that deflects the rider's path into a curve: The ...


1

It has not be proven that The Second Law of Thermodynamics is physically derived from other basic physical principles. The H-Theorem is predicated upon some pretty serious, yet plausible, assumptions about how our universe works. To my knowledge, these assumptions have not themselves been explained using other principles and/or experimental verifications of ...


3

So far, it is still an open question as to why bicycles are stable at all. There have been a few ideas put forward, but they have been disproved by construction of non-standard bicycles. The most common explanation is that the wheels on a bike act as a gyroscope, preventing the bike from falling over. A bike was constructed with counter-rotating wheels to ...



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