physics. forces and acceleration. need help w/ a problem. if you have physics knowledge and a heart, help!?
481598
There are three objects; 5kg, 9kg, and a cable of negligible mass.\n\nWhen the three move, they must have the SAME acceleration.\n\nTherefore, consider the three as a single object of mass 5 + 9 = 14 kg, with acceleration ‘a’.\n\nThe only force that is pulling the three is the force given by the 9kg object. (The gravitational pull is only on this 9kg.)\n\nThis force is 9 x 9.8 N = F.\n\nSince F = M a,   9 x 9.8 = 14 x a.\n\nFrom this a = 6.3 m/s^2.\n\nAll the three objects are moving with an acceleration 6.3 m/s^2.\n\nIf the body 5kg is moving with an acceleration of 6.3 m/s^2, then what is the force acting on it?\n\nWe know that f = ma; hence, f = 5 x 6.3 = 31.5 Newton.\n\nSince the tension of the rope is the pulling force, tension =31.5 Newton.\n\nThe tension can be calculated using the forces acting on the other end of the cable also.\n\nThe downward force on 9kg is 9 x 9.8 N; the tension is the upward force T. \n\nThe net force acting on 9kg is (9 x 9.8 - 9 x 6.3) = 9 x 3.5 = 31.5 N = T.
935341
1 sum the forces and set equal to zero. 2 forces, gravity and electric.\n\n2. conservation of energy. initially all kinetic, finally all potential. KE1=U2. U2 is electric potential. solve.\n\ngood luck
858654
the idea is that the fireman is accelerating such that ma=mg-800N, the 800N is exerted by the person holding the fireman and is facing up, the force on fireman that is pointign down is mg (a is his acceleration). where ma is mass of fireman times his acceleration. 1. find a\n2 find v, given that d=v^2/(2a) [from: d=at^2/2, v=at] where d is 10m\n3. knowing v, the ave force of impact is calculated based on impact energy and dissipation of this energy in frictional force acting for 20cm: mv^2/2=F*d, find F..., where d=20cm
985616
All are true. For the last one,\nsuppose acceleration = a.\nThen the velocity is at + c\nand the position is at^2/2 + ct + d,\nwhere c and d are constants.
210676
Use free body diagram to balance the forces and u can derive the equation of motion \nIf u want i can give u full derivation.\n"::
1093836
Centrifugal force and gravity.
883912
Let the banking angle be '$'.\nThe reaction force will be perpendicular to the banking surface.\nLet the reaction force be 'R'.\nLet 'm' be the mass of the car.\nvelocity of car (v)=27 mt/s\nradius(r)=60 mt.\n\nThe centrifugal force will be in horizontal direction. And this force will be balanced by a component of the reaction force 'R'.\nR sin$= mv^2/r ---------------------------(1) \n\nThe weight of the car will be vertically downward. This weight will be balanced by Rcos$.\nRcos$=mg  ----------------(2)\n\nNow dividing eqn(1) by eqn(2)\n\nRsin$/Rcos$=mv^2/r /mg\n=>tan$=v^2/rg = 27 x 27/60 x 9.8 = 729/588 = 1.239\n\n$= tan(inverse)(1.239)\n\nsolve above tan(inverse)  function. and find the value of the banking angle
730176
All of these questions can be answered using Newton's 2nd Law:\n\n"The rate of change of momentum of a body is proportional to the resultant force acting on the body and is in the same direction."\n\nThus, if a box has a net force of 0 N, it will not accelerate. This means that its velocity will not change. Note that this DOES NOT necessarily mean that its velocity is 0, just that the velocity is not changing.\n\nIf the box has a net force of 15 N, then it will accelerate in the direction of the applied force. Note that the magnitude of the acceleration is dependant upon the mass of the box.\n\nSo, to answer your two questions:\n\nA) Yes, it will have constant velocity.\n\nB) Yes, the box will accelerate.
910291
As long as you know mass and Force, you can solve for acceleration with the equation F = m a \n\nIf you are already learning kinetics, I'm sure you know how to solve for "a". \n\ngood luck.
391655
force = constant * distance\nF=kx\nforce = mass* acceleration\nF=ma\n\nx=F/k\nx=m*a/k = 4.0*9.8/800= about .05m, or 5cm
818860
I'm not going to give you the complete answer because you'll learn more by working through it yourself, but I'll give a few pointers.\n\nAny object travelling in a circle at a constant speed has a centripital acceleration that can be calculated from the velocity and the radius of the circle.\n\nNewton tells us that F[net] = ma, so the object must be experiencing a net centripetal force equal to its mass multiplied by its centripetal acceleration.\n\nWhere does this force come from?  In your problem, it sounds as though the only force that could be pulling the box toward the center of the circle (the centripetal force) is the friction between the box and the bed of the truck.  So, when you add up all the (horizontal) forces to get the net force, the olny one is going to be friction.  F[net] = F[friction] \n\nNow all you need to do set F[friction]  equal to the F[net] that you calculated above (remember that F[net] was equal to your centripetal force), and solve for the coefficient of friction (your textbook shold have the formula for static friction, which will include the coefficient of static friction).\n\nGood luck!
351440
well ok. for part a, the acceleration is in the horizontal direction, because its moving horizontally. so the force is just mass x acceleration which is 70*6=420N. but thatz only in the horizontal direction.\n\nfor part b, you have to account for gravity. so gravity is pulling down at 9.45m/s^2, but youre accelerating upwards at 6m/s^2. so that means that youre overcoming the gravity and accelerating at 6m/s^2 MORE than the gravitational acceleration.\nso 9.45+6 = 15.45. this is your actual acceleration.\nso now you just do F=ma again and you get 70*15.45=1081.5N. \n\nhope this helps. let me know if you need more help. and this isnt thermodynamics.
136454
I will give you a direction to solve the problem...\n\nFirst I will do the problem from a rectilinear formulation(using force, acceleration, velocity) then transfer it to a curvilinear formulation(using Torque, angular accelaration, angular velocity).\n\ngiven F=ma (or F=m(v/t) or more specifically F=m(dv/dt) or even more specifically F=d(mv)/dt).\nWhere F=force, m=mass, v=velocity, and t=time.\n\ngives you Ft=mv (conservation of energy=>conservation of momentum or the impulse-momentum equation).\n\nThis states that it will take a certain amount of force for a certian period of time to produce a certain change in velocity of an object assuming that the mass remains constant.\n\nTransmitting to curvilinear motion force becomes torque, v becomes rv(radius time velocity or w), mass becomes moment of inertia (or I) and t stays the same.
816057
again ... conservation of (mechanical only here) energy\n\npot energy = kin energy\n\nmgh = 1/2 m V^2\n\nso V = sqrt(2gh) -- see mass irrelevant\n\nV = sqrt(2*9.8*8.53) = 12.9 m/s\n\ng=9.8m/s^2 is the acceleration due to gravity\n\ng=10m/s^2 is not a bad approximation
566394
Newton's second law is \nforce = mass x acceleration\n\nHow about this? Two men were walking on a hill when they see a bar of chocolate at the bottom. Both are eager to get it. The first man, a skinny little guy of 120 lbs, trips and begins rolling down the hill. The second man, a fatty of 250 lbs, trips at the same time and begins rolling down the hill. Because the fat man has more mass, he is able to produce more force to accelerate himself down the hill faster than the skinny man. The happy fat man reaches the bottom of the hill first, and eats the chocolate.\n\nIllustrations shouldn't be too hard.
932992
it is a much easier problem than it seems..\nthe key point is the fact that the body is moving at constant speed. meaning that the net force on it is zero --> no acceleration\n\nthere are two forces. friction and the horizontal component of the force that is exerted at 40 degrees.\n\nso simply ...\n\n\nFx = Friction==>\n976 cos40 = coeff. *N *g\n\n\n now there is equilibrium at the vertical axis as well.\n\nmeaning that \n\nN+Fy = m*g =>\n\nN = (mg- fy)\n\nso we mix them together and hve\n\n976cos40 = 0.667 (mg- 976sin40) =>\nm=50.31kg\n\nin this problem you use that there is equilibrium in both of the x and y axis.
955176
Force - Newtons\nMass - Kilograms\nAcceleration - metres per second