What is 1 way u could reduce the friction between 2 surfaces?
396452
1. reduce the force at which the surfaces press together.\n2. allow one surface to slide on top of a film of air rather than the other surface(a la hovercrafts and air hockey)\n3. reduce the roughness of the surface by polishing or using lube.\n\n***I know... they're more than one way... lol***
205045
obviously you have to use the least steep slope, which will cause your marble to move slower. i think you are trying to find ways to slow down the marble itslef??? you would want to increase friction, and this can be done by applying something onto the marble or onto the slope (my favourite would be honey or syrup) im not sure if that is what you were looking for?? good luck
570511
both.  the friction, as well as creating its own heat, slows the airflow, which aggrivates the compression along the leading edges.\n\nI'm pretty sure.  don't put this on a physics test and then blame me.
344964
Use a lubricant and find a position that is comfy. Its normal, dont worry!
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...
1144673
use  lubrication use  protection
390283
coefficient of friction and specific gravity
784702
To the extent that what we call "friction" is actually primarily mechanical locking (e.g., two sheets of sandpaper, or tire rubber and concrete mutually deforming each other) there would be no effect. I think the dividing line between friction and locking is subjective. One criterion could be: if the locking parts are large enough to see it's locking, else it's friction. Or, if oiling the surfaces would let them slide freely, it's friction. On both counts, automotive braking surfaces and fan belts are frictional. \nMany man-made and geological structures would stay intact since they rely on various forms of locking for their strength. Imagine a handful of teflon-coated but rough-textured pebbles; you can still pile them up. You can pile up perfect friction-free spheres but these require an enclosure which must have tensile strength, which stone doesn't have much of. So we can have piles of rough stones but not smooth round ones. In general we could say that mountains are mostly locking structures but there could be some partial collapses. \nAdhesion would still work. (Yes, oil can ruin adhesion, but adhesion doesn't depend on a normal force so it's not frictional.) Fasteners like nuts and bolts could be modified not to unthread (e.g., lockwiring) but smooth nails would not work. Is rope-on-rope frictional? Would we lose all those wonderful variations of knots? You decide. \nOur modes of transportation would be different; traction on rough roads but not on smooth rails. Possibly rack-and-pinion gearing between locomotive and track (that's done now on steep runs). Braking could be regenerative or viscous-damped, with some sort of locking mechanism when stopped. \nWalking? Could be more like skating if shoe soles are hard and smooth and on polished surfaces. (But sometimes that's how it is even when we have friction!) Swimming, boating, flying should work fine.
496203
without friction, the car will not move when the wheel is turning. without friction, the wheel will not stop when u step on the brake. even if it stops, the car will not stop. without friction, u can't hold anything in your hand. u cannot stand upright. your glasses will not stay on your face. u will slide off your chair. everything would slide around.\n\nwithout friction, less energy is loss as heat and noise. less energy would be needed to power an engine. tire would not wear out. thing in motion would move longer.\n\nu get the idea
926088
So the first one can be solved easily using the energy method. Potential energy = mgh = 2*9.81*6\nKenetic energy  = 1/2mv^2 = 1/2*2 * v^2\nBecause there is no friction Potential Energy = Kenetic Energy:\n\n2(9.81)(6)=1/2(2)(v^2) \n\nNow you just need to solve for v, using your calculator or head.\n\nFor the second one you have to account for the amount of energy lost from friction. So, instead of KE=PE you get:\n\nKenetic Energy = Potential Energy - Energy lost by Friction\n1/2*m*v^2 = m*g*h - Friction energy \n\nSo you can do some algebra and\nm*g*h - 1/2*m*v^2 = Energy lost by friction \n\n\nThree is impossible without more information. You can't solve for u with just m, theta, and work. You need some information about the height, or the length of the ramp, or end velocity. Just think about it, you could have a ramp 100ft tall with little friction, or you could have one 1 ft tall with lots of friction. The work done would be the same even if you kept the masses and theta the same.
517904
You can call it wind resistance or skin friction.
483925
You may want to try Centripedal Force i.e. Turnings and slopes..\n\nBut it has some stuff to do with Physics as well. \n\nAlso you may want to find the maximum and minumum speeds at turnings. Include Frictional Force and try to minimize it using Calculus. \n\nTake Simple Circuits like circles, oval, parabola etc. Then take their equations and use calculus to find area under the curve etc. etc.\n\nWell thatz all comes to my mind for now!! :) BEST OF LUCK..
937470
Getting started, have to overcome static friction.  Therefore, force applied = static friction force just before movement\n\nF = umg\nF = 75N\nm = 20kg\ng = gravity\nfind u, plug and play\n\nsame problem for sliding friction
753429
Presuming you are talking about the work equation J = Fd; where J is work in Joules, F is force in Newtons, and d is distance moved in meters, then the answer is...yes you can do too much work or not enough.  Why?\n\nBecause J/F = d; which means you work to move a body a distance d.  If you were to push a mass with a fixed force F, but put in less work (J) than necessary, your desired distance moved would not be achieved.  Conversely, if you work more than required, you'd overshoot your desired distance.  Note, too much or too little work requires that a given d is specified as a required distance to move the mass.\n\nBecause you failed to describe your experiment in detail, I can only presume you are pushing various masses of various material along some surface for some specified distance (d).  Then you are calculating the amount of work done J = Fd for each mass of different material.  F in these experiments would be sliding friction force = kN; where k is the coefficient of sliding friction and N is the weight of each mass perpendicular to the surface.\n\nWhat you should discover is that the F's for each material will be different; so that, even if your d is the same for each trial, work will vary with the force exerted to slide the body.  That is, more F means more work and vice versa.  In each case, the amount of work performed is just sufficient to move the mass d meters...no more, no less.  But if you push too far, you will have put in too much work...too short, not enough work.\n\nThe F's will differ because of different k's and different N's.  In this experiment, if d is not specified, then the work done in each case is just sufficient (no more or less) than what was needed to move each different d distance.
66741
Rubber in contact with other surfaces can yield friction coefficients from 1.0 to 2.0.\n\nhttp://en.wikipedia.org/wiki/Friction
793849
i got 100N to start the container moving and 80N to keep it moving at constant speed, but i'm not sure if i got it right. What is the answer on the back of your book? if it is the same as mine, i will help you.\n\nstatic friction is how "rough" the surface is between the base of the container and the ground.\n\nnot sure about this one but i think kinetic friction is how rough between a moving object and the ground.\n\nJust think of you are stand on the hill. The static friction keep you still but once you start to slide down, the friction is less.
858739
Use Bernouillis equation and you will have your answer. You can use the equivalent lengths method to find the friction.