As a horse and wagon are accelerating from rest, the horse exerts a force of 400N on a wagon. Ilustrating?
959619
They cannot. They are acting on different bodies. The force exerted by the horse acts on the wagon, that exerted by the wagon acts on the horse. The forces are NOT colinear, which is a conditon for 'cancelling'
853525
Both the above answers are correct, provided the acceleration is horizontal. If it's not then it's a vectors and trig. question.
857622
1) I think your 2nd c.f. should say c. acceleration; that's the units you're using------ the c force of a mass m with acceleration a is f = ma, so youhave 1.5 kg*18 m/s² = 27 N\n\n2) First find the velocity: v = d/t = 2πr/t = 2π*.75m/.8sec = 5.89 m/s\nThen, acceleration: a = v²/r = 5.89²/.75 = 46.256 m/s²\nThen, the tension: T = ma = .1kg*26.256m/s² = 2.626 N\n\n3)  v = 2πr/t = 2π*90/12.3sec = 45.974m/s\na = v²/r = 45.97²/90 = 23.485 m/s²\nF = ma = 900*23.485 = 21,136.5 N\nThe lateral friction of the tires on the track
858156
The braking force was enough to dissipate the kinetic energy.\n\nForce x distance = work = kinetic energy\n\nma x 160 = mv^2/2 , notice that m cancels out\n\n160a = v^2/2, solve for v...
35590
I concur with previous answerer. The acceleration is 1.5 m/s^2. Rearrange Newton's Second Law, to solve for the acceleration. Then, substitute what you know.
193077
Force=Mass x Acceleration\n\nSo Acceleration = Force/Mass\n\nHope this helps
955845
The object has kinetic energy (1/2)*m*v^2 when first at x=0. The work friction does consumes all that energy as the spring compresses and pushes the object back to x=0. Let the distance the spring is compressed be x1. Then \n(1/2)*m*v^2 = Ff*2*x1 = mu*m*g*2*x1\n\nWhen the spring is compressed to x1, the friction has decreased the energy in the system by half and the remaining half is all stored in the spring compression because the object has come to a stop - about to start the rebound. So\nFspring*x1 = k*x1^2 = (1/4)m*v^2\nIf you solve the 1st equation for x1 and substitute the result into the 2nd, you should get the required result.
917687
Horse, carriage, feet.
832950
Well use Newtons second law. The sum of all Forces = mass times acceleration\nF = m*a\nIn the x direction the only force is friction, and the mass is the skier. The force is negative because it is slowing down the skier.\n-25 = 6*a\na = -25/6 = -4.2\n\nNow  The change in velocity is the constant acceleration times the time. \n(v - v0) = at.\n(v -v0) = -4.2 m/s^2 *20 s = -84 m/s\n\nThis is a very ridiculous problem. Most skiers are about 10 times as massive, and a speed of 84m/s is equal to 190 miles per hour.
1004891
Nope...force is what you are looking to determine.  Force equals mass x acceleration.  The fly would have to travel MUCH faster to have any impact on the train.  Also, the fly, once it made initial contact with the train, doesn't decelerate all at once...parts of it continue moving after other parts stop, and some parts start moving in East and West directions...SPLAT!
709931
kaksi_guy: I think charlesellis753 is correct.  If we interpret the question as a fill in the blank, mass appears to be the best answer.\n\nIf the ______ of an object doubles, then the force required to accelerate it at the same rate will double.\n\nThat sentence is true is the blank is filled with 'mass'.
939941
Yes you're on the right track, you know the momentum of the car along with the energy, both are conserved.  \n\nForce=Mass*Acceleration=Coefficient of Friction*Mass*Gravity\n\nYou can then solve for Acceleration=Coefficient of Friction*Gravity\nVelocity=Acceleration*t\nThen its simply a matter of filling t into this equation:\nDistance=1/2*Acceleration*t^2+Velocity*t
912569
I think F is the answer. Taking acceleration due to gravity is constant at 9.81ms^-2, but this assumes that air resistance is neglected, and g is a constant.\n\nIf air resistance is NOT neglected, and acceleration due to gravity is not taken to be constant, H is the correct answer. The rest of the options are wrong.
133048
Hello, \n\nThis problem is a bit more involved than you might initially think. There are 4 major considerations you must take into account. First, the acceleration is dependant on the angle of the hill at any given time. Therefore, acceleration is NOT constant, rather fluctuating with the incline. Second, the wind resistance is NOT constant by any means. It increases exponentially with the speed of the sled. Third, the force of the kinetic friction is not constant either. While the coefficient of friction does increase somewhat with speed (you could use an average for this), you cannot neglect the increase in friction when the sled goes around a corner. As you know, the friction is based on the normal force of the sled. At 90mph, the centripetal acceleration around a curve is V^2/r (V=velocity, r is radius of corner) which becomes quite considerable, even though steel on ice only has a kinetic coefficient between .01 - .05. \n\nNow that I have bored you out of your mind, the full equation would look something like this (I’m writing everything longhand so you know what I’m talking about)\nSum of the forces = mass x gravity x sin(angle of hill) – [(coefficient of friction) x mass x gravity x cos(angle of hill) – mass x velocity ^2 / radius of curve] – (Force of air drag)\n\nNotice the term included to account for curves: – mass x velocity ^2 / radius of curve\nDrops to zero when they are going in a straight line (r = infinity). \n\nTo find your acceleration (at a specific point in time), you simply divide everything by the mass of the sled and there you go!\n\nPS. don't take physics advice from dental students ;P
886741
It's the ground that is being pushed 'back'.  Since the ground is part of the earth, there doesn't seem to any reaction to the action of the horse pulling the cart.\n\nI'd just tell the horse that there is no reason to feed it, because in a few hours the horse would just be hungry again.
730530
Kinetic Energy, KE=1/2mv^2, where m is the mass of the person+sled equal to 77.2kg. This you got right. v is their initial velocity of 3.1m/s which you got right, too. Now substitute known values:\n\nKE=1/2*77.2*3.1^2\n     =370.9Joules\n\nThey came to rest because of the work done by the force of friction. The formula for work  is F*s, where F is the force we are interested in and s is the displacement given as equal to 30m.\n\nTherefore, Work =F*30\n\nBased on the principle of conservation of energy, kinetic energy=work.\n\n370.9=30F\nF=370.9/30\n  =12.4N\n\nCoefficient of friction=F/R where R is the total weight normal to snow which you must already have computed correctly as well.\n \nCoefficient of friction=12.4/(77.2*9.8)\n                                 =0.016\n\nLet's skin the cat another way using the formula,\n\nv^2-u^2=2as, where v is the final velocity of 0, u is the initital velocity which you correctly got , equal to 3.1m/s,  s the displacement of 30m, and a the acceleration. Now substitute known values:\n\n0-3.1^2=2*a*30\n-9.61=60a\na=-9.61/60\n  =-0.16m/s^2  it's negative because the person and sled were decelerating.\n\nNow solve for the force of friction using the formula, F=Ma.\n\nF=77.2*0.16\n  =12.4N  Same as above so coefficient will also be the same, i.e. .016, viz:\n\nCoefficient of kinetic friction=12.4/756.6\n                                            =0.016\n\n\nI also cannot figure out what I did wrong either, if you say the above solutions and answer are incorrect.
813872
The applied torque is Fapp*R=70 Nm.\nThe tension on the rope supporting the 30 kg box is m*(g+0.8) = 318.2 N. \nThe torque producing this tension is 318.2*r = 63.64 Nm.\nThe remaining torque, 6.36 Nm, accelerates the wheel at 0.8/r = 4 rad/s^2.\nThe inertia of the wheel is torque/ang. accel. = 6.36/4 = 1.59 kg-m^2.