Coefficient of Friction?
813323
Well, whilst she is sliding, ignoring air resistance, the only force acting on her is the force of friction. That's the most important part of this problem. As such, you know that Fnet = Ffriction. Ffriction = Fnormal (Uk), so you can solve for Ffriction. That's Fnet, and Fnet = ma. You know Fnet and her weight, so divide her weight by 9.8 m/s to find her mass. Solve that equation for a, and then plug it into one of the handy-dandy kinematic equations. :)
74421
centripetal force is mv^2/r.\nso it comes around 1,777,777.778 N\n\nI need one more parameter that is coeff of friction.\nif that is available u can use\n\nv(max)= sqrt [(rg (u + tan o)/(1 - u * tan o) ]\n\nassuming frictionless\n\nV= sqrt (rg tan o)\nso \no = 66.2 degrees if g=9.8\nand 65.7 degrees if g= 10.
986703
In most physics homework problems, energy is conserved, and the following is true:\n\nEo = Ef\n\nHowever, in this problem, energy is lost due to friction.  Therefore...:\n\nEo - Q = Ef\n\nWe know that the child had potential energy at the top, and kinetic energy at the bottom.  Use this information to calculate the energy lost due to friction.  The energy lost due to friction will be denoted "Q" (Q usually stands for heat, which IS thermal energy):\n\nEo - Q = Ef\n\nRearrange the expression, since we're solving for F:\n\nQ = Eo - Ef\n\nNow plug in the formulas for potential and kinetic energy:\n\nQ = mgh - (1/2)mv²\n\nYou don't have to do this, but I factored out the "m" (mass):\n\nQ = m[gh - (1/2)v²]\n\nYou can plug things in now, since the values are all given:\n\nQ = 19.6[(9.80)(3.80) - (1/2)(2.52)²]\n\nQ = 668 J
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.
181712
Friction is a theoretical coefficient used so that future results may be predicted.
515562
Friction, Heat generated to name 2 reasons.
962609
The M/A of a wedge (disregarding friction) = cot(wedge angle)\n\nUnfortunately, there's a LOT of friction, so it's not so simple after all!
726238
let mass of car = M\n\n0.6 = Force of Friction / Normal Force , Normal Force = 9.81xM\n\nForce of Friction = 5.886xM\n\nthis means that\n\ncentri force must not exceed 5.886xM\n\na=v^2 / 10, F_centri = (v^2 / 10)xM\n\n(v^2 / 10)xM=5.886xM , M's cancel out\n\nv=7.67
919106
They all involve friction, but I think you mean which has the most force due to friction, right? :-&#92;  If so, I believe the answer is that they are all the same.\n\nIn simple situations, the frictional force created by sliding two surfaces against eachother is given by:\n\nf = u * N\n\nWhere u (mu) is the frictional coefficient (an experimentally-determined constant) and N is the normal force, which is mg (mass times accelleration of gravity) if the surface isn't tilted.\n\nSo, assuming the frictional coefficient is the same for all sides of the book, there is no reason why changing the sides would make a difference.
1147561
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
948476
I won't show in details since it's hard to type out the work, but you'll get the rest, here it goes:\n\na)\nWork = mgd  *d=change in height\nW = (10kg) (9.8m/ss) (sin15) (-7.5m) = -190J\n\nb)\nW=(friction coef)(natural force)(distance traveled)\nnatural force = (cos15)(10kg)(9.8m/ss)\nW= (0.4) (cos15) (10kg) (9.8m/ss) (-7.5m) = -280J\n\nc) \nW=Force x distance\nW=(100N)(7.5m)=750J\n\nd)\n(initial KE)+ (puller work)=(gravity work) + (frictional work) + (final KE)\n(final KE) - (initial KE) = 750J - 280J - 190J = 280J\n\ne)\n(final KE) - (initial KE) = 280J\n*solve for final v\nv = squareroot (2 x [280j+ (1/2)(10kg)(1.5squared)] / 10kg) =7.63m/s
205046
the kinetic energy of the sled and person is equal to the work done by the friction to stop the sled.  Since work is force times distance in this case, the force can be calculated.  From the force you can find the coefficient from the definition of friction by dividing out the normal force.  \n\nor literally, 0.5*m*v^2 = F*d = mu*m*g. solving for mu\nmu= v^2 / (2*d*g) = 0.016\n\ni think....
355325
Newtons, or sometimes kilograms
757648
Friction force:\nF = m g c\n\nm = mass of car\ng = gravitational accel\nc = coeff. of friction\n\nCentripetal force:\nF = m v^2 / r\n\nm = mass of car\nv = velocity\nr = radius of curve\n\nSet the two forces equal to each other:\nm g c = m v^2 / r\n\nCancel m, plug in given values:\n9.8 m/s^2 * c = (12 m/s)^2 / 52 m\nc = 0.28257\n\nSo, c has to be at least 0.28257 to keep the car from slipping.
696645
Nice trick, combining multiple questions into one. I'll answer one q, no. 3.\nThe 300N force along the slope doesn't contribute to the normal force so has no effect on friction. The normal force is m*g*sin(slope) = 100 * 9.81 * 5/20. Your picture\nhttp://i34.photobucket.com/albums/d130/MATTRAINES/q4.jpg?t=1164668264\ndoesn't give a value for coefficient of friction, so if we assume it's 1, then Ff = the normal force.
462841
When the calculations were set up for determining coefficient of friction 1 was set as the upper limit. There can be no coefficient larger than 1. Highly smooth surfaces of the same shape (for example, two rectangular blocks) are in almost perfect contact with each other so that almost every molecule of one is in direct contact with every molecule of the other (that's a bit of an exaggeration unless the surfaces are micropolished, but it will serve). The extremely large surface area in contact produces enormous frictional resistance to sliding movement. It helps to understand this if you can make the distinction between smooth and slick. Think for example about trying to slide one heavy glass plate over another without any lubrication between them. They "stick" to each other like glue.
939941
Yes you're on the right track, you know the momentum of the car along with the energy, both are conserved.  \n\nForce=Mass*Acceleration=Coefficient of Friction*Mass*Gravity\n\nYou can then solve for Acceleration=Coefficient of Friction*Gravity\nVelocity=Acceleration*t\nThen its simply a matter of filling t into this equation:\nDistance=1/2*Acceleration*t^2+Velocity*t