a student weighinh 700N climbs at constant speed to the top of an 8in vertical rope in 10s. what is the power?
367171
work=forcexdistance\npower=work/time\n\n1 in=2.54cm\n8x2.54=20.32cm\n\n10cm=1m\n20.32cm=2.032m\n\nwork=700Nx2.032m\n    =1422.4Nm\n\npower=1422.4Nm/10s\n     =142.24Nm/s is the power
806029
1st link \n\nb) F = 40 N\n\nc) F = 60 N\n\n2nd link\n\nb) linear velocity = 0.59 m/s\n\nc) Taking g = 9.8 m/s^2\n\nv = 0.83 m/s\n\nd) It would not effect the answer as the expression for v which is sqrt(mu * r * g) is independent of the mass. \n\n3rd link\n\na) Rotate the setup, note the reading on the balance. Find the time required for 1 circuit.\n\nb) No work is done\n\nc) T = 5.48 N\n\nd) 5.9 %\n\ne) If the cord is exactly horizontal, it won't touch the rod and hence cannot be rotated.
461030
For the first problem, you need a law of motion. (final velocity)^2=(initial velocity)^2+2*acceleration*displacement. Substituting, you get 0=30^2+2*a*20m and a=22.5 m/s^2. Now, we need to write a force equation: ma=Tension force-weight. To find the mass of the elevator and its load, divide its weight by gravity (9.8 m/s^2) to get about 16.3 kg. And you already calculated the net acceleration. Our force equation can now be rewritten as (16.3 kg)*(22.5 m/s^2)=Tension force-160N. The tension force is about 207N.\n\nFor each of the weight scenarios in the second problem, you know the net acceleration to be 30 m/s^2. You can also find the mass of the passenger by dividing his/her weight by gravity. With the force equation ma=(force of feet up)-(weight) with respect to the passenger, you can find the force of his/her feet. Here is the work for the first scenario.\n\n16N/(9.8 m/s^2)=1.63 kg\n1.63 kg*30 m/s^2=(force of feet)-16N\nForce of feet=64.9N\n\nAnswers for the second and third scenario:\nb) 2600N\nc) 260N
438060
Without regard to air resistance you solve the problem as such.\n\nv(f) = final velocity\nv(i) = initial velocity\na = acceleration\nt = time\nx = displacement (distance)\n\nWhen you throw the ball up vertically, you know several things:\n1) v(i) = 30 m/s\n2) v(f) = 0 m/s\n3) a = -9.8 m/s^2 (this is because acceleration due to gravity goes in the opposite direction of the object)\n\nSolve for x with this equation:\nv(f)^2 = v(i)^2 + 2ax\n0^2 = (30 m/s)^2 + 2(-9.8 m/s^2)x\n0 = 900 m^2/s^2 - (19.6 m/s^2)x\n(19.6 m/s^2)x = 900 m^2/s^2\nx = (900 m^2/s^2)/(19.6 m/s^2)\nx = 45.9 m (rounded off)\n\nNow you know how high the ball went.  The time it takes to reach this is equal to the time it takes to get back down (assuming that it is the same distance up and down).\n\nTo solve for t, use this equation:\n\nv(f) = v(i) + at\n\n0 = 30 + (-9.8)t\n9.8t = 30\nt = 30/9.8\nt = 3.06 s  this is going up to max height.\n\nto check if it will be the same time going down, use this equation:\n\nx = v(i)t + (1/2)at^2\nv(i) = 0 (because when you reached max height, your velocity will be zero)\na = 9.8 because your ball is travelling in the direction of the acceleration.\n45.9 = 0 + (1/2)(9.8)t^2\nt^2 = 45.9/4.9 = 9.367\nt = 3.06 s\n\ndone
904442
The initial potential energy was m*g*h. The average speed over the run was 2000/68 m/s. If we assume the run was an incline with no change in slope, the acceleration would be constant and the final speed would be double the average.  KE = (1/2)*m*v^2. You can do the math.
905915
You're going to have to resolve this two-dimensional problem into two one-dimensional problems.  I'd suggest using vertical and horizontal.  On the vertical side, we have gravity pulling down, and the normal force and one component of the tension from the rope pulling up.  This, by the way, is why the normal force is not equal to the weight - by pulling on the rope, you're partially lifting the sled off the snow.\n\nYou know that there is no acceleration vertically, so the net force vertically must be zero.  Knowing this, you can get the normal force as a function of the tension as the weight of the sled minus the vertical component of the tension.  Knowing the normal force and the coefficient of friction will give you the force necessary to pull the sled at constant velocity.  (If the velocity of the sled is constant, what can you say about the net force horizontally?)  Since the friction is only opposed by the horizontal component of the tension, you can find that.  Once you have that, you'll find that the only variable you have left is the tension in the rope, and you can solve for that.\n\nHope that helps.
747947
Weight of bundle=4*g\nWeight component normal to the incline=\n4gcos30.\n\nForce of friction,f =uR where R is the force normal to the incline\nor 4gcos30.\n\nSubstitute given values:\nf=0.30*4gcos30\n =10.2N\n\nKE is given as equal to 128J. \nWork=f*s, where s is our unknown \n\nBased on principle of conservation of energy:\n\nKE=Work\n128=10.2s\ns=128/10.2\n  =12.5m
57888
the tension is just the force of gravity(since there is a constant velocity, the forces going up must equal the forces going down.)\nso its just the mass times gravity so 79 x 9.8
931347
x =70m, t = 7s, Vf = 15m/s\na.)\nx = 1/2(Vi + Vf)t\nVi = initial velocity\nVf = final velocity\n\n70m = 1/2(Vi + 15m/s)(7s)\n10m/s = 1/2Vi + 7.5m/s\n2.5m/s = 1/2Vi\nVi = 5m/s\nb.)\nVf = Vo + at\n15m/s = 5m/s + 7s(a)\n10m/s = 7s(a)\na = 1.43m/s/s\nbasic kinematics\nyou can't use distance = rate x time because there is acceleration and the speed is not constant.
109486
This is an Static problem:\nThe forces in the wire are:\nIn the middle of the rope the mass of Jonathan: 50kg.\nRigth side: T1\nLeft side: T2\n\nThis is a Vectorial equation: T1+T2=mg\n\nSince Johnatan is in the middle of the wire T1=T2... \n2*T1=mg\nT1 has two components T1x and T1y (horizontal and vertical)\nThe vertical component of T1 is the one that act against weight.\nT1y=T1*Sin(10)\nThen:\n2*T1*Sin(10)=mg\n[using mks system g=9.8m/s^2]\nT1=mg/[2*Sin(10)]\nT1=50*9.8/[2*Sin(10)]\nT1= 1410 Newtons.\nThe tension at "this point" is 2T1 =2820 N.\nHope it helps
840601
(a) Gravity exerts a downward force on the block of 9.8 Newtons/kg * (0.25 kg) = 2.45 Newtons. Work done by gravity = force * distance = 2.45 Newtons * 0.12 m = 0.294 Joules\n\n(b) The spring is a bit tougher since the upward force increases as the spring is compressed. Force = 250 N * (C) where C is the distance the spring has been compressed (in meters). Work is then the integral of Force over distance, so Int(250 * C (dC)) from C = 0 to 0.12 M . The integral is 125 * C^2. Evaluation at 0 gives 0. Evaluation at 0.12 gives 125 (0.12)^2 =  1.8 Joules. So the work done by the spring is -1.8 Joules . Note that the work is negative, because the object is moving against the applied force.\n\n(c) When the block strikes the spring, it will have a kinetic energy of (1/2)mv^2 = (1/2)(0.25kg)v^2 = 0.125v^2.\nThis kinetic energy is absorbed by the spring as it compresses. Note that the work done by gravity is also absorbed by the spring.\nTherefore, we have the following equation:\n0.125v^2 + 0.294 = 1.8\nv^2 = 8(1.8-0.294) = 8(1.506) = 12.048\nSpeed of block just before it hits = SQRT(12.048) = 3.47 meters/sec downward.\n\n(d) I'll leave this as an exercise for you. Just apply the same techniques as above. Note that the block hits the spring with a velocity of 2(3.47) = 6.94 meters per second...
982521
Well, when the ropes are horizontal, the child will be 2 m farther from the ground than when the ropes are vertical. (Draw a sketch if that will help.) \n\nGravitiational potential energy is m*g*h.\nYou already know that m*g is 400 N, and h is 2 m, so the increase in potential energy is 800 J.
969410
Given: m=1.6*10^7kg\n           F=8.5*10^5N\n\nI only guessed that the above is what you meant.\n\nSolve for a, using the formula:\n\nF=Ma\n\n8.5*10^5=1.6*10^7a\na=8.5*10^5/1.6*10^7\n  =5.31*10^-2m/s^2\n\nv=70km/h =70*1000/3600m/s\n  =19.44m/s\nu=0\n\nTo solve for time, t, use the formula:\n\nv=u+at\n\n19.44=0+5.31*10^-2t\nt=19.44/5.31*10^-2\n =3.66*10^2\n =366s
760545
If a weight is hanging from a V formed by two ropes, the angle of the V cannot be 180 deg, unless the weight is 0.\n\nThis is because the Tensions in the two ropes must compose to a force equal and opposite to the weight, which is acting vertically down.\n\nIf the angle is 0, then each tension is half the weight = 500/2 = 250 N\n\nFor any other angle, we need to know the relative lengths of the ropes.  If they are equal, the tensions are equal and\n2Tcos(a/2) = weight.\nwhere a is the "V" angle.
702361
ok biker bike, here we go.\n\nacc.= g = 9.8 m/s2, (any object moving in air will have this acc. or retardation)\n\nb) we apply formula\ns=ut + 1/2at^2\nu=20\nt=4\ns=20*4 + 1/2 *9.8 *4*4\n  =158.4 m\n\nc) v=u + at\n  therefore, t= v-u/a\n                   =50-20/9.8\n                   =3 seconds (approx)
83305
I suppose the problem was already due so I'll finish it for you now.\n\nF=m*a\na=F/m= 2*10^6N/1.2*10^6kg = 1.67m/s^2 * g/9.81m/s^2 = 0.17g
569320
v=vo + at\nv=0 + (9.8)(8)\nv=78.4\n\ny=1/2(vo + v)t\ny=1/2(78.4)(8)\ny=1/2(627.2)\ny=313.6m\n\nB