How can i find the percent energy lost to friction? what formula should i use?
404136
Calculate the total energy of the system before and after and make an energy balance.\n\nFor example, lets say you have an inelastic collision between two cars of equal mass m:\n\nInitial State:\nCar 1 moves at v1 to the right\nCar 2 moves at v2 to the left\n\nAssume v1 > v2\nThey hit and stick together completely.  Use conservation of momentum to find the total resultant velocity:\n\np1 + p2 = ptotal (p is the momentum of car 1 and car 2 respectively)\n\nm*v1 - m*v2 = (m+m)*vf\n\nvf is the final velocity of the clump\nthe minus sign is because I am choosing the direction "right" as positive (momentum is a vector)\n\nvf = (v1-v2)/2\n\nNow do an energy balance:\n\nEbeforecollision = 1/2*m*(v1^2) + 1/2*m*(v2)^2\nEaftercollision = 1/2*(m+m)*(vf)^2 + Efrictionlosses\n\nEbefore = Eafter (this is conservation of energy, and it is true as long as you account for all forms of energy...it wont be true if you forget about the frictional losses in Eafter)\n\nNow solve for Efrictionlosses\n\nTo get the total fractional energy loss, take the ratio Efrictionlosses/Ebefore
799229
a. Calculate the horizontal component of the 485 N.\nFh = 485 N*cos35\nCalculate the force of friction.\nFf = 0.57*319 N\nCalculate the net force.\nFn = Fh-Ff\nCalculate the mass of the box.\nW = m*g\nCalculate the acceleration.\nFn = m*a\nCalculate time to move s = 4.00 meters\ns = (1/2)*a*t^2\n\nb. Repeat with uk = 0.75
833832
Kinetic friction is equal to the Frictional force divided by the Normal force  (k = Ff / Fn)\n\nStarting at the standard equation F = ma , we need to modify this because there is some friction involved.\n\nThe two forces are our Applied force (Fa) and our Friction force (Ff).  Because Ff is acting in the opposite direction, it will be negative.  Therefore we have Fa - Ff = ma.\n\nThe key to this question is to know that because there is a constant velocity, the acceleration is equal to 0.  \n\nSo Fa - Ff = 0.  However is we re-arrange this, we have Fa = Ff\n\nThis means that the Applied force (Fa) is equal to the Frictional force (Ff).\n\nAlso, the Normal force (Fn) is equal to the Force of gravity (Fg) because we are on a level floor with no vertical displacement.  Remember though that Fg = mg.  g is yor gravitational constant.  Because there is no movement in the vertical direction, it will not be negative.  \n\nSo to solve this question, \n\nk = Ff / Fn\nk = Fa / Fg  (because like I mentioned above Ff = Fa and Fn = Fg)\n\nNow work with the number that you have.  We don't have Fg but we can solve for it using the given mass (12kg) and the constant g = 9.8\n\nk = Fa / mg\nk = 30 / (12)(9.8)\nk = 0.2551020....\n\nNoting your significant digits of 3 decimal places...\n\nThe coefficient of friction is 2.55 x 10 ^ -1\n\nHope that helps you out a little bit.
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
808463
I assume that the total travel along the incline including the spring compression is L = 3m.  The potential energy lost in this slide is m*g*h, where h = vertical distance covered.  This is L*sinø, where ø is the 35º angle.  So the total energy gained by the block and then absorbed by the spring is m*g*L*sinø.   The energy absorbed by the spring is given by E = ∫F*ds, where F is the force from the spring.  If the spring constant is k, then F=ks, and E = k∫s*ds [0 to ∆s] = .5*k*∆s^2.   ∆s is the compression of the spring.  Equate this to the  energy gained by the block to get\n\nm*g*L*sinø = .5*k*∆s^2\n\n∆s^2 = 2*[m*g*L*sinø]/k\n\n∆s = √[2*(m*g*L*sinø)/k]
703031
Well the first thing to do in a problem like this is extract information from it.\n1.) Weight of Block = 10kg\n2.) Velocity of block(Vi) = 6m/s\n3.) Final velocity of block(Vf) = 0m/s\n4.) Friction acting on block = 20N The friction is always opposite of the direction of the block.\n\nSecond:\nWhat are we calculating? Distance traveled by block! DISTANCE.  We need a formula that will give us distance as an answer! So...we have final velocity, initial velocity, and distance....lets see...how about this one...\n vf^(2) = vi^(2) + 2*a*d \n\nThis equation has Vf, Vi, and D.  \nVf = 0 m/s\nVi = 6 m/s\nD = What we are trying to find.\nThe only thing missing in this equation that we have to find is (a) or acceleration. Hmm.\n\nThere is actually acceleration on  the block.  Friction is acting on the block in the opposite direction; therefore the block is decelerating.  We can use F=ma to find the force of deceleration.\nFriction Force = 20N\nMass = 10kg\nAcceleration = ?\n\nUse the following equation:\nF=ma\n1.) -20N = 10kg * a\n2.) a = -2 m/s (Yes it's negative b/c we are decelerating!)\n\nNow we go back to the original problem!\n vf^(2) = vi^(2) + 2*a*d \n\nThus\n(0^2) = (6^2) + (2)*(-2)*(d)\n0 = 36 - 4d\n-36 = -4d.\nDistance = 9meters! (Make sure to get the units correct!)\nI think I got this right...if not tell me what I did wrong pls\n\n-----------------------------------------------------------------------------------\nThe guy who did it using Energy equations forgot to add one thing....\nmass x (velocity)^2 = Force x distance\nso 10kg * (6m/s)^2 = 20N * x\nso (10 * 36)/20 = 18 meters\n\nThis is another way to do it.  He is using the energy equations to calculate energy in the block. When the block is in motion there is Kinetic Energy in it.  When the block has no motion, there is zero Kinetic Energy in it.  However, the individual that did this messed up the equation as follows:\nKe = (1/2) * Mass * Velocity^2\n\nThus the answer is not 18 but 9 meters!!!!!\nHere is the correction:\n1.) .5*mass x (velocity)^2 = Force x distance\n2.) .5*10kg*(6m/s)^2 = 20N * x distance\n3.) (5 * 36)/20 = distance&#92;\n4.) Distance =9 meters!
826674
use conservation of angular momentum in these kind of problems\nNOT energy conservation\nif you dont know how to use the conservation equation, better read up on examples
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.
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
822266
Conservation of energy:  KE=PE\n1/2*Mb*Vb^2= (Mb+Mw)*g*h\nVb^2=2*(Mb+Mw)*g*h/Mb\nVb^2=2*(0.01+2)*10*0.1/0.01\nVb^2=402\nVb=20.05 m/s
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
891305
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)
1024834
compression and friction
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
544891
The formula is actually divided up into two cases:\nForce friction = Force Applied when Force Applied < mu Static x Force Normal (i.e. nothing'll move below this threshold)\nForce friction = mu dynamic x Force Normal otherwise.\n\nThere is no formula for the two mus; they are measured experimentally.
711024
Look, task 1,3,4 are very alike. We shall use the same formulas there:\n1. let us find energy of the body at A, at B, and at C\nat A:   mV^2/2 + mgH = m(V^2/2 + gH)\nat B H=0, so the energy constitutes the kinetic energy mV(b)^2/2\nat C: mV(c)^2/2 +mgH(c) = m(V(c)^2/2 + gH(c))\nthe energy does not change (not dissipates), so energy in A equals the energy in C:\n\nV^2/2 + gH = V(c)^2/2 + gH(C), so \nV(c) = sq.root {V(a)^2 + 2g(H(a) - H(c))} = \n = sq.root{64 - 2x2x9.81} = 4.976 m/s\n2. The force of gravity for the mass 100 kg will be \nF = mg = 100kg x 9.81 m/s^2 = 981N\nAs the mass is moved along steep surface, some part of the gravity force will hinder the movement. We can fin the component of the gravity force along the surface using the triangular 20m and 5 m proportion, (because the triangles are similar), so\nFg(along) = 981N x 5/20 = 981N/4 = 245,25N\nWe have to apply the force 300N to move the body along the surface, hence we overcome the gravity component 245.25, and the rest is for friction.  \nSo the Ffriction = F - Fg(along) = 300N - 245.25N = 54.75N\n\n3. The formulars are as in problem 1. \nat the start   Es = Ekinetic + mgH\nat the highest (final) point of the bounced ball\nEf = Ekineticf + mgHf,\nHere we have to make some remarks (very important)\nIf the question would be "How high the ball can bounce",\nthen we could assume that the highest point would be when the ball bounces vertically up, and at its highest point it has zero velocity, no vertical one, no horizontal one. Than we can makethe energies equal (assuming no dissipation) and find Hf:\nmV^2/2 + mgH = mgHf\nso,  Hf = H + V^2/2g = 4(m) + 36(m^2/s^2)/2x9.81(m/s^2) = \n             = 4(m) + 1.835(m) = 5.835 (m)\nIn case the ball bounces not vertically, but along the steep line, at his highest point it will have zero velocity too, so the answer will apply to the case. \n\n4. The velocity in B will be found using the formula for energy at starting piont and at B:\nmV^2/2 + mgH = mV(b)^2/2,\n\nV(b) = sq.root{V^2 + 2gH} = sq.root{16 + 12x9.81} = 11.56 (m/s)
1133901
Use the equation U = 0.5*k*x^2\n\nU = potential energy (joules)\nx = compression distance (meters)\nk = force constant (N/m)\n\nThis gives us the potential energy of the toy.\n\nU = 0.5 * 296 * 0.0122^2 = 0.0220 J\n\nThen calculate the Kinetic Energy:\n\nKE = 0.5 * m * V^2\n\nKE = Kinetic Energy (joules)\nm = Mass (kg)\nV = Velocity (m/s)\n\nKE = 0.5 * 0.153 * 0.299 = 0.0229 J\n\nTotal Energy = 0.0229 + 0.0220 = 0.0449 J\n\nFor part 2, assume total energy has been converted to potential energy and solve for compression distance:\nx = sqrt (2 * U / k) = sqrt (2 * 0.0449 / 296) = 0.0174 m\n\nAmplitude is twice the compression distance (for positive and negative)\nAmplitude = 2 * 0.0174 = 0.0348 meters\n\nFor part 3, assume total energy has been converted to Kinetic energy and solve for speed:\nV = sqrt (2 * KE / m) = sqrt (2 * 0.0449 / 0.153) = 0.766 m/s