A 0.30kg ladle sliding on a horizontal frictionless surface is attached?
916915
1.Kinetic energy at equilibrium point\n=1/2*m*v^2\n=1/2*.3.v^2\n=10J\nv^2=66.66\nv=8.16 m/sec\nAs the ladle is moving at this speed\nthe spring will come in to action\nHowever at the equilibrium position \nthe spring tension=zero\nHence at the instant the ladle is\n passing the equilibrium \nposition work done =zero\n2. At .1m tension force in the spring\n=500*.1=50N\nWe assume the velocity remains the same\n(strictly velocity will decrease due to deceleration\nimposed by the spring)\nWork done =50N*8.16 m/sec\n=408 N-m/sec\n=408watts
835350
The coefficient of friction is the ratio between the force pressing two objects together and the force of friction between them. In the case of of sliding friction, this is literally the ratio of two forces. However in the case of static friction, this is the ratio of one force (the force pressing the two objects together, also called the normal force) to the maximum force that can be supplied by static friction. The actual force may be less than that. For instance, if the force tangent to the surface is less than the coefficient of friction times the normal force, the actual force of friction will be equal and opposite to the tangent force, and the object won't move.\n\nAnyway, the question asks whether there are any materials for which the coefficient of friction is greater than 1. A simple glance at a reference table, such as the one linked below, reveals the answer is yes.
65267
It's me again, Dhinesh,your classmate.\nrolling frictin is the friction caused due to rolling. it opposes rolling & it acts on a single point as there is a single pt. which has contact with surface
898058
Its the ratio of the lengths.\n\nIts balance of angular momentum, so the length from the hinge and the forces exerted must be equal.\n\n10 N exerted upward at  2 cm from the fulcrum means that the downward load 2mm the other direction from the fulcrum is associated with a load of 100 N.\n\nYou may not have noticed, but there are special force concentrating deformations in the surface near where the tab presses.  That makes it fail at about 1/2 to 1/3 the load it would otherwise fail at.
1022513
You probably mean "identify the frictional force acting on this object." Since we don't know what the object is, we can't really help you...
110772
0.58 (with a horizontal bar right above the 58)
878371
The incline is a simple machine that permits less force to be applied to move an object. The work associated with moving an object on an inclined surface, in the absence of friction, is the same as raising the object of weight (m*g) through a distance H.\n\nFor the inclined plane with an inclination angle of 30 degrees, moving the mass (m) over a distance L would raise the object a distance H given by\n\nH = L*sin(30) = 0.5*L\n\nThe work then is\n\nW = m*g*H = m*g*L/2\n\nFor the values given\n\nW = (1.0 kg)*(10 m/s^2)*(0.5 m)/2\n\nW = 10/4 N-m = 2.5 N-m\n\nThe same answer can be found by looking at the net forces on the object and multiplying by the distance traveled.\n\nFu = 10 N (the upslope force)\nFd = m*g*sin(30) = (1 kg)*(10 m/s^2)/2 = 5 N (downslope force)\n\nFn = Fu - Fd = 5 N (Net force)\n\nW = Fn*L = (5 N)*(0.5 m) = 2.5 N-m
69030
Initially, the ball is pushed, it slides. There is no rotation, as stated in the problem/question. This means there is no rotation at all. That all the velocity is translational.\n\nFor this scenario think of the ball as a solid block that has been pushed and it sliding across the floor.\n\nAs stared, at some time (t1) the ball begins to roll.\n\nAt this time, the ball's motion is both rotational as well as translational. In other words, it rolls as well as moving forward. This motion is what takes place at the *contact point* of the ball to the floor.\n\nIf the ball rolls, it is not slipping.\n\nIndeed, the friction causes the linear motion to decrease, and eventually stop the ball.\n\nThe rotational motion of the ball is also undergoing friction. A rolling sphere's motion is very subtle and often misunderstood. By virtue of the ball's mass there is a very tiny (tiny is relative to the mass and material) flat spot between the ball and the surface upon which it rolls, whereby a small torque gives rise to a resistance of the motion. This is why a rolling sphere slows. But.....since there is friction, the velocity of translation, is related to that of rotation by the following:\n\nv^2 = 2gh/(1 + I/Mr^2)\n\nWhere M is the mass of the ball and "r" its radius, I its moment of inertia. This is common notation. If there is no rotation, then the moment of inertia term drops out (I = 0) and you are left with a velocity which is strictly that of a translating body......kind of like free-fall.\n\nIf the ball is given one initial push to give it the initial velocity Vo, then unless there is some other forcing function at work, friction will act to decrease the motion of the ball, and all said velocities thereof. None of the velocities will increase.\n\nIf you know of a situation where such a case can exist where the ball increases velocity without doing anything to it.....make sure you mention my name when you pick up the your Nobel Prize.\n\n:)
44667
You have a contradiction in your statement.  Static friction implies no movement but velocity is movement.  \n\nIf you mean the velocity of an object in dynamic equilibrium has a result of a frictional force (kinetic though) like sliding a block across the floor then you have to know the amount of work put into the system.  Then use work-kinetic energy theorem.
497010
DUE TO SURFACE TENSION POSITIVE ANGLE OF CONTACT AND INTERNAL PRESSURE.
792649
If it were on a frictionless surface, the ball (or any object) would remain stationary relative to the ground and slide backward relative to the wagon, until it fell off the wagon (assuming there's nothing preventing that). But since the object is a ball, I'm guessing the question involves the ball rolling, which will happen if there is friction. In that case the ball won't remain stationary since it is being accelerated in rotation, and the tangential force needed to do that, applied through frictional contact, also accelerates the ball forward. The ball will move forward relative to the ground, but backward relative to the wagon.\nSo not enough information has been supplied for a definite answer.
761342
We did something like this in my physics lab with an Atwood machine.  It was a pully with weights hanging from it.  I can't exactly remember how it worked, but maybe this will inspire you.  We also did something with an air track and these two masses colliding and we'd see which one would go in which diretcion.  You need a nearly frictionless surface, like an air hockey table or ice slab.
848443
when u will draw ur inclined plane and mass on it there r different forces acting in it \nfirst force will be directly perpendicular to the ground level i.e. Fg=mg\nsecond force acting is mgcos(theta) which is perpendicular to hypotenuse of the inclined and makes up Fn=mgcos(theta)\nthen another component parallel to the hyp. of the inclined \nwhich equals mgsin(theta) which goes down the ramp\nand  then u can\nequal the force going down with the frictional force\nmgsin(theta)-(mew)mgcos(theta)=ma\n(as we know frictional force is (mew)Fn=(mew)mgcos(theta) )\nsolve by putting values\nmew=0.146    ans
53108
PSEUDO WORK DONE\nThis is because of the pseudo work done by the inclined surface or plane.\nExplanation:- Let theta be the angle of inclination of the inclined smooth plane and h be its height.Think off an object(body) of weight W=mg is sliding down on the plane. The another force that is acting on the body is force offered by the plane against the weight of the body, let this be R, this force is acting perpendicularly to the inclined plane.Further due to the interaction of forces W & R there exists a resultant force by which the body is sliding down,let it be P,this force is acting along the line of inclination.Now in view of lami`s theorem we have, P=W sin(theta)= mg sin(theta);whence theta is the angle of inclination and 0<or equal to theta <or equal to pi/2.Hence it can be noticed here that sin(theta)is <or equal to 1.\nTherefore P<W ie. mg ( weight of the body).This clearly implies that force less than the weight of the body is enough to slide up to the height h through the inclined plane.For instance when Theta=0 inclined plane will be horizontal,consequently P=0.And when theta = pi/2 ; P will be = mg, this implies that the body falls freely.As a typical example let theta= pi/6.01...,P is< 1/2 mg.Therefore the force -P=-1/2 mg orP=1/2 mg is enough to slide the body to the height h.Accordingly work done by the force while lifting the body to h through inclined plane = 1/2 mgh \nMoreover, as h is the height of the inclined plane,Potential Energy (gravitation potential)at h is = mgh,which is also equal to the work done while lifting the body to the height h, by any path.ie. P.E=mgh. (P.E stands for potential energy).Here true work done is =mgh; work done by P = 1/2 mgh \nie. mgh-1/2mgh=1/2 mgh,This shows that there is a shortage of 1/2 mgh of P.E, when work is done through inclined plane.How to account for this shortage. Actually, there is no bargain in the laws of nature.\nTo account for the above shortage please once again consider lami`s theorem.\nie. R=W cos(theta)= mg cos(theta);where R is the force exerted by the plane on the body, perpendicular to the inclined plane.When theta is pi/6.01, R=1/2 mgh (nearly)\nTherefore pseudo work done by the plane = 1/2mg (nearly)\nie. work done by p + work done by R = True work done\n     1/2mg + 1/2mgh = mgh\nNote-plane is considered as frictionless.\nI think this is enough to understand the phenomenon.
783057
What I said about inclined surfaces works just the same on a horizontal surface; you will compute sin(theta)=0 and cos(theta)=1 so the normal gravity force=g*m and there is no gravity force parallel to the surface.\nWhen KE or PE decreases due to friction energy loss, the decrease equals the friction energy loss which is friction force * distance. I assumed you already knew that. If my answer helped I'm happy. I'd appreciate it if you go back to that question and select my answer as best; this is the normal procedure even though it would eventually be selected by public vote.\nP.S. A tip on yahoo answers. I only found this question by accident. You can add followup questions to your original question; just click on the original question and there should be a link that lets you add more words, probably called Additional Details. That's what I was referring to when I said I'd be checking for followup questions.
704245
the constant velocity indicates that the 30.0 N force is just enough to overcome the force of friction\n\nSo use 30 N in the friction formula:\n\nmu=Force of Friction / Normal Force , Normal Force = Weight of Crate\n\nmu = 30.0/(9.81 x 12 ) = 0.2548419979612640163098878695209
932992
it is a much easier problem than it seems..\nthe key point is the fact that the body is moving at constant speed. meaning that the net force on it is zero --> no acceleration\n\nthere are two forces. friction and the horizontal component of the force that is exerted at 40 degrees.\n\nso simply ...\n\n\nFx = Friction==>\n976 cos40 = coeff. *N *g\n\n\n now there is equilibrium at the vertical axis as well.\n\nmeaning that \n\nN+Fy = m*g =>\n\nN = (mg- fy)\n\nso we mix them together and hve\n\n976cos40 = 0.667 (mg- 976sin40) =>\nm=50.31kg\n\nin this problem you use that there is equilibrium in both of the x and y axis.