Airplane: what is coefficient of kinetic friction for this problem? Or what are the steps to answer this?
931515
Seems like a bogus question. The wheels can't avoid slipping in the first part of that .53 seconds. Do I misunderstand the question?
851078
1. Usually you have to solve these problems in terms of energy, where initial energy = final energy + friction (lost) energy. The initial and final energies are each the sum of the respective initial and final potential + kinetic energies. (Sometimes the problem doesn't explicitly give you an initial potential energy; it may just tell you that an object slides x meters downhill.) If you know the initial and/or final momentum and either the mass or the velocity, you can find the missing variable (velocity or mass) and then transform the momentum into kinetic energy. \n2. The coefficient of friction 'mu' of course affects the friction force. Sometimes you are asked to find mu so you have to work backwards. You may know the lost energy and the length over which the object encountered friction. Then you solve for friction force. Then, knowing the normal force as well (force perpendicular to the friction surface), you can find mu.\n3. Inclines with friction are a big favorite of the problem writers. Just remember that the gravitational force normal to the incline is g*m*cos(theta) and the gravitational force along the incline is g*m*sin(theta). If a ramp is tilted just enough to allow the object to slide, g*m*sin(theta) = friction force = mu*g*m*cos(theta), so mu = sin(theta)/cos(theta) = tan(theta). Or if you know mu and want to find theta, it's arctan(mu).\n4. Another fave is centripetal force on a horizontal turntable with a friction surface. We know the centripetal force is w^2*r*m (w being angular rate in rad/sec). And we know the normal force is g*m. So, if we rev up the turntable until the object slides off, or we move the object out from the center until it slides off, we know that, at the instant it lets go, w^2*r=g*mu, so we can find w or r. Easy, right?\nI hope this helps; if you have followup questions I'll be watching this question until it closes.
873659
First one's pretty easy.  Since you have a frictionless track (and we're assuming there is no air resistance here, either), the speed is due solely to the change in potential and kinetic energies.  At the start, the cube has a potential energy of mgh, or 0.530 kgX0.387mX9,8 m/s^2 = 2.01 J.  By the time it reaches the bottom of the loop, it has converted all of that potential energy into kinetic energy (since we're calling the bottom of the track to be zero altitude).  As it travels up the loop, it gains potential energy again as it fights against gravity, slowing it down.  At the top of the track, therefore, it has a potential energy of 0.30 J, meaning the kinetic energy must be the difference, or 1.71 J.  Solving for velocity (KE = 1/2 mv^2), you get 2.54 m/s.\n\nThe second case is a little more tricky, in that it involves rolling friction.  However, friction only plays a role in starting and stopping motion -- if the cylinder achieves pure rolling motion, the frictional component goes to zero because there will be no relative motion between the points of contact on the ramp and the cylinder (no sliding).  If we assume that there is never any slipping along its path, then it should behave just like the cube sliding along a frictionless track, and the speed will still be 2.54 m/s.  I could be wrong, but I think that's the answer.
397135
The weight of the object is irrelevant unless one is taking into account air resistance. This is because the net force on the object is the force of gravity, so we have ma = -mg, or a = -g, which is all we need to know for the vertical kinematics equations. The only thing we need to know for the horizontal equation is the initial horizontal speed of the package.\nIf, on the other hand, one was concerned with air resistance, one would also need to know approximate air density and cross-sectional area of the object, as well as be able to numerically solve a nonlinear ordinary differential equation (the frictional force is proportional to some function of velocity) so I doubt this is what you wanted.
939533
I would use conservation of energy.\nx is the distance traveled on the ramp\nthe forces at work are:\nGravity.  The potential energy will be\nsin(5)*m*g*x\n\nFriction:\n.48*cos(5)*m*g*x\n\nSince there are no brakes all of the kinetic energy will be lost to friction and the potential energy gain:\n\n.5*m*v^2=\n.48*cos(5)*m*g*x+sin(5)*m*g*x\nnote that m divides out\nI get 110 m\n\nj
724326
ASSUME V IS OWN SPEED OF THE PLANE & W IS WIND SPEED\n(V+W)*2=225  - WHEN FLYING WITH WIND\n(V-W)*3.75=225 – WHEN FLYING AGAINST\n2 EQUATIONS!!!
1039676
Newtons second law - blance of momentum.\n\nWhen the force acting on the mass to accelerate it is the maximum that the engine, atmosphere, and gravitational attraction to the earth can produce, and those forces are balanced by the the forces acting on the mass to slow it down, then the mass is at its "max speed".\n\nDensity of the air is affected by altitude, temperature, humidity, and weather.  For horizontal, noncompressible flight, that density is a major component of the "dynamic pressure" and therefore the drag.\n\nA very simple model of drag is D=0.5*rho*U^2 *Cd*A where rho is the density of the working fluid (atmosphere), U is the freestream speed of the air, Cd is the coefficient of drag of the airplane, and A is the frontal cross-sectional Area.\n\nIf you set max thrust equals this drag, then you can math out what U is, and call it the max speed of the plane.  If the plane has a headwind or tailwind then its absolute speed will vary.\n\nIf the plane approaches mach then stranger and more energy-demanding phenomena kick in.  If the plan is going beyond the speed of sound there is a third, also complex description for the drag.\n\nIf you are using a prop driven plane then you cant just assume that the thrust equals the planes horsepower, you have to consider the efficiency of the prop and there are equations for that.\n\nDetermining the Cd for a plane is complex because the wing, body, and tail all interact to change the overall Cd.
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
882711
a)\nRate = Distance/time = (24000 ft)/(20 min) = 1200 ft/min\n\nb)\nTo write A(t), you know that the max. value is 24000 and the plane descends at a constant, linear rate of 1200 ft/min.  Thus, you can write a linear equation of this model:\n\nA(t) = 24000 - 1200t, where t = time (in minutes)\n\nPlugging in 0 (at initial time), the height is 24000 feet, which matches the fact in the problem.\n\nc)\nThough linear functions have infinite ranges and domains, this problem is limited to the realism of the problem.  The range is only limited from the maximium value (24000 ft) to the ground (0 ft).  You can't have a negative value for altitude because that would not make sense in the problem.  The domain is limited to the time in air (20 min).  You can't have a negative time and time after 20 min is irrelevant because the plane has already landed.  Summarizing:\n\nDomain: 0 <= t <= 20\nRange: 0 <= A(t) <= 24000\n\n--------\n\nHope this helps
699073
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
910713
at the moment there are the horizontal component of weight and the friction that opposes it are in equilibrium. we must apply a force that with the already present weight compontns will equal the kinetic friction. \nmeaning that \n\nWx + Fp = Friction.\n\n500cos20 + Fp = 0.6*500sin20 =>\n\nfp =367.24N
456850
Do your home work. \n\nStill I tempt to answer this question.\n\nW= weight of the box\nmu= coefficient of friction.\n\nWe assume there is no acceleration. So force to upward movement is Fup = Wcos(20)*mu + Wsin(20),\n\nsimilarly Fdown = Wcos(20)*mu - wsin(20),\n\nso\nWcos(20)*mu + wsin(20) = 4Wcos(20)*mu - 4wsin(20)\n5Wsin(20) = 3Wcos(20)*mu\nmu = 5sin(20)/3cos(20)\nmu= 5/3 tan (20)\nmu = 0.606
758142
a) There is an equation derived from the motion in x and y that results like:\n\ny = (tanθ )x - g x²/ [2(Vo cosθ)² ] we know y, x and θ we need to solve for Vo:\n\nVo =  (x / cosθ) √[ g / (2 (x tanθ  - y )]\nIf we put  the plane at the initial hit at 1 m, y is 21 m -1m = 20 m\nAnd as we know that x = 130 m and θ= 35°\nSubstituting:\n\nVo = 41.7 m/s <<<<<<<<\n\nb) t = x / ( Vo cosθ )  Substituting: t = 3.81 s <<<<<<<\n\nc) Vfy = Vo senθ - gt   (Vfy is the final velocity of y after the time t)\nSubstituting, its components are:\nVfy = -13.46 m/s      <<<\nVox = Vfx = Vo cosθ = 34.16 m/s   <<<\n\nand its magnitude:\n\nVf = √(Vfx² + Vfy²) = 36.71 m/s <<<<<
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.
858739
Use Bernouillis equation and you will have your answer. You can use the equivalent lengths method to find the friction.
878213
f = ma = kN = kmg; so a = kg = .15*9.81; where a is the deceleration, k = .15 the friction coefficient and g = 9.81 m/sec^2 the acceleration due to gravity at Earth's surface.\n\nWork = fd = kNd = 1/2 m v^2 = kmgd; where friction force f = kN times distance (d) is work (energy in Joules), which in this case comes from kinetic energy 1/2 m v^2 where v = 5 m/sec and m is the mass of my damn physics book.  Upon rearranging factors to find d, we have:\n\nd = 1/2 v^2/(kg) = v^2/(2kg); where d is the stopping distance, v = 5 m/sec, k = .15 the friction coefficient, and g = 9.81 m/sec^2 from gravity.\n\nYou can do the arithmetic.  The important physics lesson learned here is that neither answer (a or d) depends on the mass of that textbook...and both depend in large measure on k in that larger k means greater deceleration and smaller k means longer stop distance.
734736
first one is easy\n\nfirst find the normal force\n\nFf (friction) = Fn(normal force ) (.35)\n0(motionless ) = Fn .35\nFn = -.35\n\nfrom there its easy\n\nFn = mgsin(theta)\n\nsolve for theta\n\n-.35 = (200)(9.8) sin(theta)\n\nand solve it ur self\n\nsecond one, tell us wat u want us to find and we can tell u, but here are the things u can find, ill give u the formulas\n\nacceleration = 0 (constant speed)\ntheta = 15 degrees\nvelocity = constant\n\nthats pretty much it, from what u gave me