PHYSICS QUESTIONS- ENERGY PROBLEMS (DIAGRAMS INCLUDeD!)?
827544
1) use conservation of energy:\n\n1/2*m*vb^2=1/2*m*va^2+m*g*h\n\nm=2\nva=4\ng=9.8\nh=6\nv=sqrt(16+2*9.8*6)\n=11.56 m/s\n\n2.\nSame concept\nm*g*(bounce)=1/2*m*va^2+m*g*(drop)\nmass divides out\nbounce=va^2/(2*g)+drop\nva=6\ng=9.8\ndrop=4\nbounce=36/(2*9.8) + 4\n=5.84 m\n\n3)  since the mass isn't moving it's offset the 300N force\n\n4)\nAgain, use conservation of energy\nNote that the change in potential energy will subtract from the kinetic energy of the start \n\n1/2*m*vc^2=1/2*m*va^2-m*g*delta-h\nthe mass divides out\nva=8\ndelta-h=2\ng=9.8\nvc=sqrt(64-2*9.8*2)\n=4.98\n\nj
757859
haha, k  Draw a picture and draw a free-body diagram.\n\nI don't feel like busting out a paper/pencil  so I'll try it in my mind\n\nF(sub g) cos Theta  =  Mg cos(20 degrees)\n\nbecause the Normal force is equal to the y-component of gravity.\n\nTwo similar triangles, the incline and one which has the Force of gravity straight down, and the y compentn equal opposite of the normal force, and X component equal opposite to F\n\n\nlots of physics i'm skipping in my explanation tho
865948
KE = 1/2 mv^2\nfind the KE before the collision using the formula.\nfind the KE after the collision using the formula.\nsubtract the two values to find the change in kinetic energy
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
735831
Did you put your entire homework assignment on here? I noticed you posted a bunch of science questions just now.\nIf you don't have your textbook handy [you really should be looking up these answers so that you can remember and understand them when you have your test/exam], go to the links below and look them up. The 1st is a down loadable physics textbook. The 2nd has animations of physics.
885634
using conservation of energy the kinetic energy at the bottom is \n1/*m*v^2\nThe frictional force times the distance is the work done by friction\nand the displacement in the vertical direction is the potential energy that gets converted to kinetic and friction\n\n1/2*m*v^2+f*L=m*g*h\nwhere h /L= sin(15.9)\n\nPlug in the numbers and solve using algebra\n(.5*6500*3.8*3.8)+4300*L=(6500*9.8*sin(15.9))*L\n46930+4300L=17451*L\n46930/(17451-4300)=L\n=3.6m\n\nj
915417
Basically, in conservation of energy, you have:\n\nEnergy(initial) = Energy(final).  This becomes\n\nKE(initial) + PE(initial) = KE(final) + PE(final), neglecting air resistance.  When the projectile hits the ground, there is no potential energy because h = 0.  Therefore, the equation simplifies to:\n\nKE(i) + PE(i) = KE(f)\n(1/2)m*v_i^2 + m*g*h = (1/2)m*v_f^2.\n\nWe know that:\n\ng = 9.8 m/s^2\nv_i = 175*sin(45) m/s (because we are only taking the vertical component)\nh = 165 m\n\nThe masses cancel, now solve for v_f:\n\nv_f = sqrt([v_i*sin(45)]^2 + 2gh) = sqrt(175^2 + 2*9.8*165)\n\nSolving, you get v_f = 184 m/s\n\n------------\n\nHope this helps
754118
E1=M*g*h, M=(your mb)=32kg, - pot energy of the system at rest,\nh=2m and g=9.8m/s^2.\nE2=kinetic of M=32kg\nE3=kinetic of m=20kg\nE4=kinetic of pulley u=5kg\nE5=potential energy of m at the height h, E5=m*g*h\nThus energy conservation: E1=E2+E3+E4+E5.\nE2+E3=(M+m)*V^2/2, V to be found.\nE4=I*W^2/2, angular speed being W=V/r, r=0.1m, and I=u*r^2/2 for your pulley.\nHence 2*M*g*h=(M+m)*V^2+(u*r^2/2)*(V*r)^2+2*m*g*h,\n(M+m+u/2)*V^2=2*(M-m)*g*h, - does not depend on r.\nV=2.08m/s, don’t believe & check it! – good picture gal – a piece of art.
443144
you should try this site http://www.oberlin.edu/physics/dstyer/SolvingProblems.html
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
463421
KE = mv\nWrong: energy <> momentum\n\n2y = (2 m)cos(kx)\nRight: a lenght =  lenghth x a number between -1 and +1\n\nvf = vi + ax\nWrong: x should be t\n\ndensity = volume/mass\nWrong: fraction should be inverse
734494
(a) Work done by gravity = change in potential energy of the book. Now change in potential energy = m*g*h2 - m*g*h1=2*10*(10-1.5) =  170J. So, work done by gravity is 170J\n(b)Change in gravitational potential energy = work done = 170J as explained above.\n(c) Potential energy when textbook is released = m*g*h1 = 2*10*10=200J\n(d) Potential energy when it reaches the hands = m*g*h2 = 2*10*1.5=30J\n(e) If U =100J at ground level, work done by gravity = 170J, the work done does not change with change in reference.\n(f) change in potential energy also remains 170J\n(g) U at release point = 100 + 2*10*10 = 300J\n(h) U at hands = 100 + 2*10*1.5 = 130J
735311
the potential energy of the ball:\nPE = mgh = (.6 kg)(g)(.7m)\nwhen the ball collides with the block all the Potential Energy has been converted to Kinetic Energy so,\nPE = KE = (.6 kg)(g)(.7m) = (1/2)(m)(v^2)        (v^2 is v squared)\nall that kinetic energy is transfered to the block so\nKE(block) = 4.12 Newtons\n\nin one dimension the solution is that all momentum and kinetic energy is transfered to the block such that,\n(a) ball speed after collision = 0 m/s\nand \n(b) block speed after collision\n(1/2)(m)(v^2) = 4.12 Newtons\n(1/2)(2.5 kg)(v^2) = 4.12 Newtons\nv = 1.8 m/s
746965
You don't need to find the mass if use conservation of energy to solve the problem.\n\nThe kinetic energy of the can initially is:\n1/2*m*v^2\n=.5*m*2^2\n=2*m\nand when it comes to rest it is zero\nThe energy was lost to the work done by friction.\n\nthe force of friction times displacement is the work:\n\nu*g*m*d\n=.2*9.8*m*d\n=19.6*m*d\nnow set the energy of the system equal to zero:\n2*m-19.6*m*d=0\ndivide out the mass and solve for d:\nd=19.6/2\n=9.8m\n\nj
771697
Try Google or Yahoo.  You'll find worked examples there.  I think it needs diagrams to understand; they and the accompanying explanations are too complex to attempt to reproduce here.
971746
This problem involves kinetic energy, force of kinetic friction, work, and conservation of energy.\n\nGiven: weight of girl + sled=645N\n           mass of girl +sled=645/g\n                                        =6.58N\n           speed of girl +sled at bottom=7m/s\n           coefficient of kinetic friction=0.050\n\nKinetic Energy, KE at the bottom of the hill=1/2mv^2\nKE=1/2*(6.58)7^2\n     =161.2 Joules\n\nForce of kinetic friction, f=uR, where u=coefficient of kinetic friction, and R is the force normal to the surface equal to weight of girl+sled.\n\nf=0.050*645\n =32.25N\n\nWork=f*s, where f is the force of kinetic friction and s is the distance the sled travels. Thus\n\nWork=32.25s\n\nBased on principle of conservation of energy:\n\nKE=Work\n161.2=32.25s\ns=161.2/32.25\n  =5.0m
611013
Oh good grief.  If it's anything like what I've just done it'll include velocity, centrepetal forces, magnetic fields, friction, gravity, forces generally, working out the mass of something and the speed at which it will fall, balances with pivot points, static electricity, AC & DC electricity, transformers, vectors and so on and so on.  It's quite hard but it's maths based so if you like maths, you should be ok.  I wish you luck.