Physics: Inclined Plane?
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.
869908
the force of friction is N*mu, where N is projection of mg onto the axis perpendicular to the plane surface, so N = cos(20deg)*mg,\nF_friction is then N*mu=cos(20deg)*m*g*mu, this force is opposing the motion and is parallel to the plane.\nthe other force is projection of mg onto the axis parallel to the plane wnad it is sin(20)mg, so the total force is F= cos(20deg)*m*g*mu+sin(20)mg\n\ndraw a diagram to get the projection right, ie using cos or sine. I find it helps to visualize the angle of inclination becoming small and looking at what angle in the projection vs axis is gettings smaller
962881
if acceleration is on the x axis, the slope represents 1/mass\n\nas\n\nslope = x/y = acc/force\nand\nforce=mass x acc\nso\nacc/force=1/mass\n\nif force is on the x axis, the slope represents mass\n\nas\n\nslope = x/y = force/acc\nand\nforce=mass x acc\nso\nforce/acc=mass
473777
Wedge is a movable inclined plane; Some times two inclined plane  are joned toghther.Hence it is a simple machine,\n\nIn inclined plane, the plane is fixed and load is moved upward by effort.\n\nIn wedge load is fixed and effort and the plane are moving and hence do work.
817519
hey ya.\nso, most of this question is irrelevant static.\nthe main point is, there's a patient on a tilted surface, and it might slide off. \n\nso if the patient isn't sliding off, what's keeping him from sliding down off the bed and onto the floor?\n\nif the answer is friction, then you are correct.\nif you don't think the answer is friction, what else could it be? alternatively, if the patient wasn't on a bed, but rather, on a greasy slide, it wouldn't take much of an angle for the patient to slide off.\n\nokay.\nso it's friction. \n\nyour book probably has done the old "friction between a slope with angle A and a box with mass M sitting on it" question to death. \n\nthe deal is that if you can't pass through a surface (so it's not made of water or jello or something), then it'll push back on you with a force pointing STRAIGHT out of the surface. this force is called the NORMAL FORCE. how much force does it push out with? the amplitude of the normal force is JUST ENOUGH to counteract the force you (or gravity) push down on it with. so. gravity will pull you down into it, and the normal force will push you back. \n\nanyway.... this is better explained with diagrams. look it up in your textbook. it's the old "incline plane" problem.
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..
800064
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
871196
Don't you just need to balance forces in orthogonal directions and then use these components to get the forces along the cable?\n\nSince the sign is stationary, vector sum of forces must be zero.  Sketch a free body diagram and label known forces.
723693
I'm pretty sure you are correct.  I am also working thru Calc III (at the moment, actually).  It helps to picture the gradient as perpendicular to the tangent plane.
111071
I think you are referring to the coefficient of static friction u(pronounced as myoo)s. This involves some derivation of equations but if you'll review your static friction on inclined planes, you can easily start from there.
35729
Polarized in a vertical plane
943006
The balance of forces acting on you: m*a=f1-f2; m is your mass, f1 is the force pulling you down the slope, f2 is force of friction, both being parallel to the slope incline. Thus f1=m*g*sin(b), f2=m*g*cos(b)*k, while g=9.8m/s^2, b=31deg, a=1.16m/s^2, m*g being your weight. Mind! While f1 is a vector parallel to the slope, the force m*g*cos(b) is normal to the slope and should not be taken into account unless friction be present. \nNow m*a=m*g*(sin(b)-k*cos(b)), hence k=(sinb-a/g)/cosb=0.46
1110486
It comes from the fact that the "hypotenuse" of the two "triangles" is not a straight line segment.  It is actually two segments of similar, but not equal, slope.
716765
The potential energy at the top is mgR = 2 x10 x R = 20 R.\n\nKinetic energy at the bottom of the curve = 0.5 m U ^2 = 0.5 x 2 x U^2. = U^2.\n\nAs there is no friction  K.E = P.E\n\nU^2 = 20 R.\n\nOn the horizontal path, \n\nInitial speed is U.\n\nFinal speed is V = 0.\n\nDistance traveled = 8 m.\n\nAcceleration = Fricttional Force / mass = -8/ 2 = - 4 m/s^2. ( Negative sign since the force acts opposite to the motion.\n\nUsing V^2 - U^2 = 2as, and also noting that V=0,\n\nU^2 = 2 x 4 x 8 = 64.\n\nU = 8m/s.\n\nThe time of travel is found using V= U+ at.\n\nt = -8/ (-4) = 2 second.\n\nUsing U^2 = 20 R.\n\nR = 64/ 20 = 3.2 m.
783086
Gravity is pulling down on the object, air is resisting it (pushing up, if you will) and causing friction, and if there's any wind it's pushing sideways on the object. I doubt you'd have wind in your problem, though. Wind is complicated :)
770005
correct. It would be parallel to the x-axis and therefore have a slope of zero. Versus the above equation which has an undefined slope.
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.