New answers tagged

-1

While applying work energy theorem you missed the work done by friction:-So,$$\frac{1}{2}\cdot m\cdot v^2 - \frac{1}{2} \cdot m \cdot 6^2 \neq 0-0$$But,$$\frac{1}{2}\cdot m\cdot v^2 - \frac{1}{2} \cdot m \cdot 6^2 = W_{friction}$$ $W_{friction}$ will not be zero because it always opposes the relative motion between the inclined surface and the body.In fact ...


0

Work is $\int Fds$ so what matters here is the direction of the displacement AND the force. Yes the total displacement is zero but you have to consider that the friction force changes direction when the motion changes direction, and thus when dividing it into two stages, the first stage the friction would be acting down while the object is being displaced up ...


-2

Weegee, this is a two step problem, for which I will give the general approach and leave the details to you. On the way up the incline, an equation needs to be set up that accounts for all energy terms. With "KE" equaling kinetic energy, "PE" equaling potential energy, and "$W_f$" equaling work done by friction, the equation is: $KE_{bottom} + PE_{bottom} ...


-3

If there was no friction then the body would rise so far up the slope, stop, and slide back down, returning with the same speed it started out with. But with friction, it will be slowed down both on the way up and on the way down. The greater the friction the greater the effect. So you definitely do need to take friction into account. You also need to ...


0

I don't see how the power delivered to the wheels gets to be kinetic energy of the car without friction doing work. Forces don't have to do work. Imagine a compressed spring with a mass. The compressed spring has a bit of energy in it. When we release the spring, it can accelerate (do work) on the mass. This seems pretty simple. But for the mass to ...


0

The wheels exert a backwards force on the ground, so the ground exerts a forward force on the wheels. That's what actio = reactio tells us. Since the center of the wheel is (largely) stationary relative to the car, the impulse (force times time) delivered from the ground must be forwarded at the axis of the wheel. Here we have again the axis of the wheel ...


0

Work has two different but equivalent definitions. One is that it is a force applied over a distance, and the other is that work is a transfer of energy. I prefer the second definition, as it is a bit clearer in circumstances like this. The car accelerates, and therefore there is clearly an increase in KE. But was work done by the road? The road is ...


1

The friction just provides the grip. The torque of the engine does the work. Its the same when you start running- your legs exert the force and friction just prevents your feet slipping backwards.


0

Well, I would compare the velocity of each block relative to the other blocks, for the bottom block, I would compare its velocity to the block above it, and if it is moving with respect to the bottom block, then there must be kinetic friction there. The max static friction applied for the top and bottom of block 1 would be different, though, because there is ...


0

Yes. Take your example of positive work. The reason that the amount of work done on the block is positive is that the force on the block is in the same direction as the block's motion. But the force on the belt is in the opposite direction of the belt's motion, and therefore the work done on the belt is negative.


0

If the belt is decelerating then static friction will be opposite to the motion of object and will do negative work.


-1

Work done by all dissipative forces are always negative, including kinetic friction, because they always reduces kinetic energy. When you place object on moving belt - it starts to move in opposite direction to belt speed in a belt reference frame. So belt induces kinetic friction until that object comes to rest in belt reference frame. So the same $𝑊<0$ ...


1

Yes work done by kinetic friction may be zero for example:- consider a block slipping on ground work done by kinetic friction will be negative in ground frame but now observe the block w.r.t block itself now work done by each and every force will be zero as displacement of block w.r.t itself is zero.


-1

Kinetic friction is a dissipative force opposes relative motion. A "kinetic frictional force" that does no work is non-dissipative and not friction at all. Such a force has absolutely no impact on the velocity of the object. Kinetic friction always does work for any non-zero distance.


3

Hold a piece of wood against a sanding belt. In your frame, the block is not moving, but kinetic friction is exerting a force: you have to hold the block still energy is transferred: the block gets hot, and pieces are pulled off it


0

Generally, no- the effort required to lift an object from a surface is not materially increased by the factors that cause resistant to motion across the surface. However, when two very flat and clean surfaces are pressed together the forces of attraction between them can exceed the effects of gravity. To understand why this is the case you should consider ...


0

What is your purpose in doing this experiment? It shows that objects in motion tend to stay in motion, while objects at rest tend to stay at rest. The quarter demonstrates that by not flying around the room randomly. And also by not missing the cup when a small sideways force is applied. It sounds like you want to minimize the sideways motion, and you ...


0

The mass doesn't make a difference. The two competing effects represented by your bullet points cancel each other out, so you will get the same result whether the coin is a heavier or lighter one.


0

ma = mgf where f is the coefficient of friction. As you can see the mass of the coin cancels out. So there are 2 things here that you can control. First is the coefficient of friction so the smoother the surfaces the more ideal. 2nd is the acceleration on the card itself so in comparison the acceleration of the coin would be negligible.


1

I can't really answer this. But I think the concrete road will have more sharp edges that make tiny cuts in the tires, and the tar will be a little bit sticky and pull at them in multiple directions. The tar might tend to slightly dissolve in the tire and vice versa. Would that make the tire wear faster or slower? It would surely reduce miles-per-gallon, ...


0

Yes. There is a maximum acceleration that your tires can provide.Be careful with sharp turns at high speeds where the centripetal acceleration is close to this limit. Any additional acceleration (or breaking) may cause the vechicle to slip out of the turn. If you decrease your speed, centripetal acceleration decreases and you will be more safely within ...


0

Yes, everything you say is correct assuming, as we do in general, that dynamic friction (kinetic friction) is pretty much constant. If you are interested in more accuracy than that general assumption provides, then we can talk about how in reality dynamic fiction does slightly vary with speed, typically decreasing with speed depending on the materials ...


1

It has to do with this: In a perfect world, turns would all be banked so you could traverse each at full speed and not skid sideways off the road. Since it is not always possible to bank the roadway surface at the right angle to support a full-speed turn, the reduced speed limit in a turn is set to accommodate the existing bank angle and turn radius minus a ...


1

As outlined by OP in a comment, we are dealing with an object lying on a surface with one side fully in contact with that surface. Kinetic friction doesn't depend on area. In only depends on normal force (the pressure on the surface):* $$f_k=\mu_k n$$ $\mu_k$ is the friction coefficient and can be thought of as a constant (depends on the two materials, on ...


0

Normally the friction will not do any work during a quick collision of two hard bodies, like billiard balls. Nevertheless, it will have observable consequences in the form of rotation: part of the initial kinetic energy will be converted into rotational kinetic energy. I think if you assume perfect conservation of energy you might even be able to calculate ...


0

With friction operating in the system between the two bodies while impact(elastic),mechanical energy conservation law still holds good because in elastic impact friction doesn't do work as it is instantaneous(somewhat impulsive). Linear momentum may be conserved if there is no external force on two bodies but as friction(between colliding bodies) will be an ...


0

When there is friction $f$ acting on a ball of mass $M$ and radius $R$, the friction delivers both linear impulse and angular impulse to the ball. In a time step $\Delta t$, the change in linear velocity is $$ \Delta v = \frac{1}{m}f\Delta t $$ whereas the change in angular velocity is $$ \Delta \omega = \frac{1}{I}fR \Delta t = \frac{5f\Delta t }{2MR}$$ ...


0

In the ground frame of reference we have a single force acting on body. This is the normal force from the side wall of the groove. This force always acts tangentially so it accelerates the body tangentially so as to keep it in the groove as the table rotates. However, there is no radial centripetal force to keep the body moving in a circle. So in the ground ...


3

The groove applies a normal reaction on the block, which is the reason that the block rotates along with the table with the same angular velocity as the table. Now since this normal force is tangential to the table it causes the tangential speed of the block to increase. Tangential velocity is equal to angular velocity cross radius and angular velocity ...


0

The body would keep its state, as no force is acting on it (first law): if the body is at rest with respect to the ground frame, it would remain at rest; if it has some speed, it would keep that speed. However, as seeing from the rotating reference frame, in both of these cases the body is accelerating. So, in order to describe the motion from that non-...


1

If the groove is sufficiently deep/steep-sided to retain it, and assuming the body starts off somewhere in the groove that is not at the exact centre of rotation, then the body will slide to the end of the groove. That will happen because the wall of the groove will press against the body and accelerate it and there is no centripetal frictional force to make ...


2

The friction force opposes the direction of the movement, not necessarily the weight's parallel component. If the tension force is greater than the weight's parallel force the object will want to slide up the ramp and the friction force will be pointing opposite of the tension force along the ramp. The static friction force will try to hold the object ...


1

The thing confusing me is which direction will be the force ma acting on the body? There are only two forces on the mass. There is gravity, and there is the normal force from the ramp. The sum of those forces will generate a resultant acceleration. Some force should be acting on the plane giving it acceleration a then at the same time body will also ...


2

The thing confusing me is which direction will be the force $ma$ acting on the body? If the body is at rest with respect to the incline, then it must be only moving (accelerating) horizontally. By Newton's second law, this means that the net force acting on the object is horizontal to the right with no vertical component. Some force should be acting on ...


0

The work is done by gravity. Friction simply holds the instantaneous contact point stationary to the ramp, so it doesn't do any work. The rotation is around the contact point, and is cause by gravity acting through the centre of mass of the roller.


0

Here it is more appropriate to say that the net work done by friction is zero, because friction here support rotation and at the same time opposes translation motion.so amount of work done by friction in rotation is cancelled out by work done in translation.So the loss of potential energy is equal to the Gain in total kinetic energy.and here friction acting ...


1

You are basically describing a solenoid engine, these are quite functional, and could probably be made large enough to propel a full sized car. Although not a very energy efficient, or smooth operating design for an electric motor. Here is a video link to a small 4 solenoid engine: https://www.youtube.com/watch?v=x4im3M9IFcI If you type in "solenoid engine" ...


0

You have some of the ideas right. Magnetic forces are what makes an electric motor run. But things are usually arranged differently. You are suggesting using an electromagnet to push and pull a magnetic piston. Then the car's crankshaft will convert back and forth motion to rotary motion. It is more common to arrange magnets and electromagnets so they ...


1

For a frictional force (i.e. one that slows down the particle), $b$ must be negative. When you set $b<0$, you obtain: $$x(t)=\frac{i|b|^{1/2}}{(mv_0)^{1/2}}\tan\left(\frac{(mv_0)^{3/2}}{i|b|^{1/2}}t\right)=\frac{|b|^{1/2}}{(mv_0)^{1/2}}\tanh\left(\frac{(mv_0)^{3/2}}{|b|^{1/2}}t\right)$$ since $\tan\left(\frac{k}{i}t\right)=\frac{1}{i}\tanh(kt)$. This ...


1

It is the static friction between the bottom of the tire and the road surface that keeps the tire from slipping or skidding (kinetic friction). Even though the tire is rotating, the bottom of a non slipping tire is in static contact with the road. When applied force is greater than static friction the wheel will slip. This can be caused by hard braking (...


0

. . . . at what point does the wheel stop spinning and stop causing a skid? If the wheels are not rotating and the car is moving then the car will be in a skid. The condition that there is no relative movement at the point of contact between the road and the tyre is that $v=r \omega$ where $v$ is the speed of the wheel axles, $r$ is the radius of the ...


0

You are applying a constant force, but the constant force happens to exactly equal that from kinetic friction, hence the total force, which sets the acceleration the block experiences, is zero and it moves with constant speed, equal to that of your hand. You see, this kind of experiment is one that, because of its reliance on a human participant - an active ...


0

It is just because the table is (1) giving enough friction to slow down the object so that it is always touching your hand, but (2) your hand can push with greater force than the friction can act on the object. To see for (1), imagine if the table is very smooth and you put some soap water on it, and now push a metal ball. Another scenario is to have a lot ...


0

The object seems to always travel at the same velocity as my hand, does this mean I am not actually applying a constant force? Yes it does mean you are not applying a constant force. The force is due to an interaction at the surfaces of your hand and the object, and that interaction depends on how closely in contact and compressed those two surfaces are. ...


0

It's very difficult to apply a constant force to a small object with your hand. Your hand will move at the speed your brain commands it, rather than applying a constant force. Try pushing something very heavy, such as a boat, car, or a trolley stacked with drinks, stones or another person. Then you will have to push with constant force (the max force you ...


0

Other answers have correctly explained that it's very difficult to apply a constant force by hand on an object on a table. However, if we change a bit the experimental setting we can apply a fairly constant force on an object by hand and feel it accelerating. Part of the problem of pushing an small object on a table is that the forces involved are small and ...


0

Experiment: Try moving your mouse (if you use one) across the screen at a constant speed. This should feel difficult to achieve, especially if you aren't resting your wrist or elbow on anything. Your brain wants the mouse to move at a constant speed, but your muscles are probably overcorrecting every time you see the cursor speed up or slow down. The ...


2

I actually think this is kind of a great question, but I'd like to focus on one of your statements in particular: "The object seems to always travel at the same velocity as my hand, does this mean I am not actually applying a constant force?" If you are pushing an object, this will be true regardless of the force you apply. That's part of the definition of ...


3

The maximum force that friction can apply is given by the equation $F = μ N$. Here, $N$ is the normal force, which on a flat rigid surface is equal in magnitude to the weight of an object. The coefficient of friction is $μ$, which is related to how much "grip" the surface has - how rough or smooth it is. Ice has a low coefficient of friction, while sandpaper ...


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