a student wants to determine the kinetic or sliding friction coefficient beteween a box and a plank.She places a box on the plank and gradually raises one end end of the plank. when angle of inclination with the horizontal is 30 degree. the box starts to slip and it slides 2.5m down the plank in 4.0s. what is the kinetic or sliding coefficient of friction?
205932
sorry dear friend u would need homework help site
245428
Average Kinetic coefficient of friction (m) cannot be calculated as such, \n      since  m can lie anywhere between 0 and 1.\n   \n      Thus, its value is probably more near 0.5 than many values. That's all that can be infered.
342415
40N on a frictionless surface would accelerate 5 kg at 8 m/s² (F = ma) so there must be a total  of only 6.666N acting on the ice. That means that a frictional force of 3.333N is acting in the opposite direction.  Since the coefficient of kinetic friction is calculated as the frictional force divided by the normal force, it must be\nFk = 3.333/5*9.8 = 3.333/49 = .068\n\n\nDoug
416477
For these problems, it would be helpful to draw a free-body diagram to represent all the forces on this problem.  Here you have 4:\n\n1) Weight in the vertical direction going down, let's denote it W\n2) Force, moving in the x-direction, denote it F_1\n3) frictional force, moving opposite of Force, denote it F_2\n4) Normal force, perpendicular place of contact (opposite direction of W), denote it N\n\nNow use Newton's 2nd law to write equations in each direction:\n\ny-direction: \nN - W =0 (because the block does not move in the y-direction).  Therefore:\n\nN = W, where W = m*g\n\nx-direction:\nF_1 - F_2 = 0 [because the block is just about to move when F_1 is applied].  Therefore:\n\nF_1 = F_2, where F_2 = u*N\n\nMaking the necessary substitutions for F_1 and N, you get:\n\n30 = u*(4.4)(9.8).  Solving for u, you get --------->\n\nu = 0.7\n\n--------\n\nHope this helps
348740
(2) You need to consider conservation of energy. The energy of the ball at point a is\n\nm*g*Ha + 0.5*m*Va^2\n\nwhich, neglecting losses, is completely converted to kinetic energy at point b. Equating them gives\n\n0.5*m*Vb^2 = m*g*Ha + 0.5*m*Va^2\n\nand solving for Vb\n\nVb = sqrt(2*g*Ha + Va^2)\n\n(3) Same considerations, however when the ball reaches its maximum height (Hm) all of the energy is potential.\n\nm*g*Ha + 0.5*m*Va^2 = m*g*Hm\n\nSolving for Hm\n\nHm = Ha + 0.5*Va^2/g\n\n(4) If the block is moving at a constant speed up the ramp due to the force F then summing the forces gives\n\nF - m*g*sin(q) - Fr = 0\n\nwhere q is the angle of the ramp, m is the mass of the block, g is the gravitational acceleration, and Fr is the friction force.\n\nSolving for Fr\n\nFr = F - m*g*sin(q)\n\n(5) Again, conservation of energy, but this time at point c the ball has both kinetic and potential energy.\n\nm*g*Ha + 0.5*m*Va^2 = m*g*Hc + 0.5*m*Vc^2\n\nSolcing for Vc\n\nVc = sqrt(2*g*(Ha - Hc) + Va^2)
438081
a=( change in velocity)/(change in time)\n  =(Vfinal - Vinitial)/2\n  = (5.5-11.2)/5 \n  =--1.14 m/s2\n\nForce of Friction = mxa \n\nsince we don't know the mass of the puck\nlet m=M\n\nForce of Friction = mxa = 1.14*M\n\nCoefficient of Friction = Force of Fiction / Weight\n        \nWeight = mxg  again we don't know puck mass m=M\n \nWeight = 9.81x M\n\nPlug this into\n\nCoefficient of Friction = Force of Fiction / Weight\n\n=1.14xM /(9.81xM) , the M's cancel out\n=1.14/9.81\n=0.116
493517
The frictional force acting against the motion is F\n\n= (mu)x mg x cos a. \n\n= 0.2 x 22 x 9.8 x cos 25 = 39 N. \n\nThe net force acting along the inclined plane\n\n= (mg sin a +39) Since both the forces are acting opposite to the motion.\n\n= 22 x 9.8 x sin 25 + 39\n\n= 130. 12 N. \n\nIf x is the distance moved along the plane, the work done by this net force is Fx.\n\nThis work done is used to remove the kinetic energy of the body = 0.5 m v v.\n\nFx = 0.5 m v v \n\nx = 0.5 x 22 x 3.9 x 3.9 / 130.12 = 1.28m.
362613
Well we have\n\nwavelength = h/momentum = h/mv = h/m(.0131c)\n\nwhere h is plank's constant and c is speed of light.\nso substitute and calculate.
484248
In the absence of friction, the acceleration of an object down a 12-degree slide would be\n\n9.81 * sin 12, or around 2.04\n\nInstead, your obejct is accelerating at 1.22.\n\nThe loss of acceleration due to friction is\n\n(9.81 * sin 12) - 1.22\n\nand the coefficient of friction is thus\n\n[(9.81 * sin 12) - 1.22] / (9.81 * sin 12)
403865
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.
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Consider the torque of the system. The weight of the plank, acting at the centerof mass, is a distance d to one side of the pivot and produces a torque given by\n\nTp = d*Wp\n\nwhere Wp is the weight of the plank and d is given by \n\nd = L/2 - y\n\nwhere L is the length of the plank and y is the amount by which it overhangs the pivot. The walker, on the opposite side of the pivot, produces a torque\n\nTw = x*Ww\n\nwhere x is the distance from the pivot and Ww is the weight of the walker.\n\nWhen the torque of the plank is greater than the torque of the walker the plank will remain horizontal across both supports. However, when the torque of the walker exceeds the torque of the plank it will start to rise. The furthest the person can walk before the plank starts to tip can be found by equating the two opposite torques.\n\nTw = x*Ww = (L/2 - y)*Wp = Tp\n\nSolving for x\n\nx = (L/2 - y)*Wp/Ww
551220
Sliding friction is the friction that an object has while it's moving.\n\nAn eraser has the most sliding friction.\n\nStatic friction is the resistance of an object from movement. I suppose since an eraser has the most sliding friction, it will also have the most static friction. that or a paperback book just because it's heavier.\n\nit really depends on the coefficient of friction. \n\nI should say that the eraser will have the highest coefficient of frictions, but the paperback book will have the highest overall friction just because of it's weight.
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Net force on object = Force applied - Friction force\n\nSince velocity is constant, the acceleration of the object is zero\n\nSo, Net force = mass * acceleration = zero\n\n0 = Force applied - Friction force\nor\nForce applied = friction force \n= coefficient of friction * Normal force\n=.250 * 11.9 kg * 9.8 m/s^2\n=29.2 kg-m/s^2\n\nBox is decelerating at .250*9.8m/s^2\n\nand\n\ndistance traveled = (final velocity^2 - initial velocity^2)/2a\n=(0 - 3.2 m/s)  /  2*(-.250*9.8m/s^2)\n=.65 m
488598
you need to work out the force of the book thats heading down the ramp using triginometry. as its a constant speed, there is no acceleration, so the friction must be equal to the force pointing down the slope
372375
Well, according to me F= (force of friction)(normal force) is the equation to be used.\n\nSo the force exerted in the x- direction is 30N and the normal force is (12kg * 9.8)N\n\nThus you get when you solve for the force of friction\n\n(30N) / (117.6N) = .2551\n\nSo the friction between the crate and the floor is .2551. \n\nThere are no units because the newtons cancel out when you divide.
470790
First, decompose the forces due to gravity on the carton:\n\nPerpendicular = cos(22)*m*g\nParallel = sin(22)*m*g\n\nAssuming the carton is sliding, compute the force due to friction:\n\nf=Cos(22)*m*g*uk\n\nso the acceleration of the carton is\n(sin(22)-Cos(22)*.12)*m*g/m\nor (sin(22)-Cos(22)*.12)*9.8\n=2.58m/s^2\n\nTo calculate the velocity\naverage velocity is 1/2a*t\nand distance is\n1/2a*t^2\nwe know that the distance is 9.3m\nso we can compute t as\nsqrt(9.3*2/2.58)\n= 2.68s\nthe velocity at the end is\n2.68*2.58\n=6.9m/s\n\nj
553884
For 4 a&b)\nThere would be three arrows.  The first would be a force in the upward direction (i.e. normal force) second would be force due to gravity(downward direction, with equal magnitude o that of the normal force).  The third arrow is an arrow that is in the same direction as the force applied to the large box.  Its magnitude is proportional to the normal force.  It s static friction that is applying that force.   \n\nAs far as part c goes It's been a while for me.  You've got the textbook, look up some formulas.  Besides, I think you'd need to include some masses in order for me to solve those.\n\nThe force diagram for the Large box is very similar to that of the small box, except that it has forces applied to it from the small box. (this is because the large box is supplying the normal force for the small box.)