What is the total weight force of a woman's foot slamming on the brakes of a car??
464267
weight of human, speed of car, amount of torque applied to break pedal for working service, even the shoe type would be a factor.....all in all there isn't any way to apply a formula due to unimaginable variances. but the weight of her boyfriends foot slamming into her butt for not watching the road...that's much easier..ha ha
861868
what's the 30 g for? here's my answer:\n\nfrom F=m*a,\n\nF= (70 kg)*(3m/s^2)\nF=210 newtons...\n\nthis is the force needed to help the 70 kg person survive the crash...does that help?
539746
F = ma\n\nF - Force\nm - mass\na - acceleration
726238
let mass of car = M\n\n0.6 = Force of Friction / Normal Force , Normal Force = 9.81xM\n\nForce of Friction = 5.886xM\n\nthis means that\n\ncentri force must not exceed 5.886xM\n\na=v^2 / 10, F_centri = (v^2 / 10)xM\n\n(v^2 / 10)xM=5.886xM , M's cancel out\n\nv=7.67
311499
Work = force * distance\n= 1000 N * 10 m\n= 10000 Nm\n\nYou wouldn't take into consideration the mass of the car because the force you are applying is horizontal (assuming that you are pushing straight on the car and not down at an angle).  The only time you would use the 3000 N is if you took friction into consideration.  In that case, you would have a friction force opposing your 1000 N, thus reducing it.  (Force of friction = coefficient of friction * normal force)\n\nPower = work / time\n= 10000 Nm / 100 s\n= 100 Nm/s
757156
How about someone is run over by a car.  The owner of the car says they left it in neutral on a level parking lot, and must have forgot to set the parking brake -- the car started rolling forward before they could jump back in and stop it, and the car ran the victim over.\n\nThis can be disproven because Newton's first law (the "Law of Inertia") states that an object at rest will stay at rest until acted on by an outside force.  Since no outside force was present (the parking lot was level so no gravitational force acts to accellerate the car horizontally) and the car itself was turned off, it cannot possibly have "just started to roll by itself" -- the car owner intentionally ran the victim over.\n\nWhy would he do that?  Well I guess that will be the fun and imaginitive part of the assignment!
884398
The work done on the car by the net force equals the change on the car's kinetic energy.\n\nWork = Force * distance\nKE = 1/2 mv^2\n\nWe are given a mass for the car and the final velocity, so from this we can calculate the car's final KE,\nKE = 1/2 (2000 kg) * (2 m/s)^2\nKE = 4000 Joules\n\nSince in this case, KE = W\nW = 4000 Joules = F * d\n\nWe know the net force to be the sum of all the forces acting on th car.  Since there is a "forward" force of 1140 Newtons and a "backward" (what we will call 'negative') force acting on the car of 950 N, the net force is 1140 - 950 = 190 Newtons\n\nW = 4000 J = (190 N) * d\nd = 4000 J / 190 N\nd = 21.05 meters
757648
Friction force:\nF = m g c\n\nm = mass of car\ng = gravitational accel\nc = coeff. of friction\n\nCentripetal force:\nF = m v^2 / r\n\nm = mass of car\nv = velocity\nr = radius of curve\n\nSet the two forces equal to each other:\nm g c = m v^2 / r\n\nCancel m, plug in given values:\n9.8 m/s^2 * c = (12 m/s)^2 / 52 m\nc = 0.28257\n\nSo, c has to be at least 0.28257 to keep the car from slipping.
1155452
Only impulse (force x duration, or the quantity of momentum transferred to the pin) is calculable from the data, not force by itself, which is variable in time and not directly meaningful.\n\nIf the moving object has mass M and velocity V, and the object being struck has mass m, then assuming the collision is elastic and all motion is in the forward direction (all these assumptions are at least approximately true for the bowling ball and pin) then the impulse is:\n\nImpulse = 2*M*m*V/(M+m)\n\nEach pound value must be divided by g to get proper units of mass, so this is:\n\n=(g/g^2)*[2*16*3.5/19.5]*V = 5.74*(V/g)\n\nSince:\n\nV/g = 23mph/(32ft/sec^2) = 1.05 seconds \n\nthe result is:\n\n6 pound-seconds\n\nIt's the equivalent of 6 pounds acting for 1 second, or 1 pound acting for 6 seconds, or etc..
198734
The force of friction is dependent on two things, the Normal force between the "object" and the "surface" and the coefficient of friction between the two.\n\nThe normal force on a level surface is equal the weight of the object.  You are given the mass, from this, you can easily calculate the force of weight (and this F_n).\n\nYou are given the coefficient of friction, greek letter mu.\n\nThe force of friction can be found as,\nF_f = F_n * mu\n\nThe force of friction is always in the direction opposite that of the motion.  So if your traveling "forward", the frictional force is pointed "backward".\nWhat are the two components in the formula that affect the magnitude of the frictional force?  Just the normal force (which is dependent on mass and gravity) and the coefficient of friction.  For all intensive purposes, mu is constant, as is the mass and gravity of the object.  Therefore, the force of friction is constant.\n\nYou are given in initial velocity.  You are trying to find how far (change in distance) the car will travel while this constant force is being exerted on it before the car comes to a stop.\nThere are several ways this could be found.  I think the energy method would be the easiest, but I will cover the kinematic method as well....\nThe car has some initial kinetic energy when the problem starts out.  KE = 1/2 mv^2.  After the car comes to a stop, the car has zero KE...therefore some work had to be done on the system to change the energy (we assume a level road).  The change in KE equals the work done.  Work is also equal to the product of the force applied over the distancd applied.\nW = F * d\nSince the frictional force is what is causing this change in energy (which we know, it was calculated earlier), and we also know the change in KE, we can simply solve for the distance, d.\n\nThere is also another possible solution.\nWe know that the frictional force will cause an acceleration.\nF = m*a\nWe know that an acceleration will cause a change in velocity.  We know ,\nv_f^2 = v_i^2 + 2ad\nwhere v is velocity (i is initial, f is final), a is acceration, and d is distance.  Simple solve for d.
704333
let us say that she is walking in the x direction (positive) \n\nThis is a question of momentum... \np = mv   \n(momentum equals mass times velocity)\n\nand we have the conservation of momentum \np_1  +p_2 = p_(total)\nThe sum of the momentums is equal to the total momentum of the system\n\nSo in the begining she is not moving... so the total momentum of the system is zero... it will always be zero.  \n\nShe walks... her momentum is p_1 = 45*1.5 = 67.5 kg*m/s\n\nand we know that p_1 + p_2 = 0\n\nso \np_1 = -p_2 \nor \n67.5 = - m v = - 150 *v\nso the velocity of the plank is -0.45 m/s.  \n\nNow we just sum the velocities to get the total velocity and we see that she is traveling at 1.05 m/s relitive to the ice. \n\nGood question ... i have not thought about a problem like this in a long time
973163
Relative to a woman in the car:\n\n(a) the translational velocity is ZERO since the woman is moving at the same translational velocity as the car relative to the ground.\n\n(b) the magnitude of the translational acceleration of the center of the wheel is also ZERO because the center of the wheel is NOT moving and NOT rotating relative to the woman.\n\n(c) v, velocity of the point at the top of the tire is = 80 km/h because that is how fast the car is moving, thus the wheel must also be rotating at the same speed.\n\n(d) a, acceleration of the point at the top of the tire is given by:\n\na = v^2/r = (22.22)^2/(0.33) = 1,496.45 m/s^2\n\n(e) v, velocity of the point at the bottom of the tire is = -80 km/h because at the bottom of the tire, the tire is moving in the opposite direction\n\n(f) a, acceleration of the point at the bottom of the tire is the same as (d) a = v^2/r = 1,496.45 m/s^2\n\nNow relative to a hitch hiker on the ground:\n\n(g) v at the center of the wheel is = 80 km/h or 22.22 m/s, since the car and the center of the wheel (which is not moving relative to the car) are moving at the same speed relative to the ground.\n\n(h) a at the center of the wheel is = v^2/r = 1,496.45 m/s^2, since the center of the wheel is rotating about the bottom point of the tire as the tire rolls on the ground.\n\n(i) v at the top of the tire is v(top) = ωd, where ω is the angular velocity of the rotating tire and d is the diameter of the wheel. The reason we use diameter of the wheel is because the top of the tire is rotating abou the bottom of the tire, not the center.\n\nω = v/r = (22.22)/(0.33) = 67.33 rad/sec\nv(top) = ωd = (67.33)(0.66) = 44.44 m/s\n\n(j) a at the top of the tire is = v(top)^2/d\n    = (44.44)^2/(0.66) = 2,992.29 m/s^2  \n\n(k) v at the bottom of the tire is equal to ZERO because the bottom of the tire is not moving relativ to the hitch hiker or the ground.\n\n(l) a at the bottom of the tire is also equal to ZERO since the velocity is zero.
40451
well figure out the net force with F=ma and just subtract. like just say 93 x .26 is the net force, the force thats acually being used for motion. so you subtract this from 36 to see what force is opposing this. so the total equation is:\n36 - (93 x .26)
193077
Force=Mass x Acceleration\n\nSo Acceleration = Force/Mass\n\nHope this helps
19306
You see martial artist break very thick boards all the time and they probably can't lift a car. It is about centering your strength. Besides a car weighs more than 300 lbs.
545558
Force = mass * acceleration\nForce = 1300 kg * 2.6 m/s2\nForce = 3380 Newtons
111044
I'm just doing all your work aren't I?  Its ok I like physics, and its good practice.  I am studying for the MCAT coming up this April (Med School).  This problem unlike the other ones that I did is a little more tricky to me.. Mostly because I don't have my physics textbooks with me.  But it would appear that you need to calculate the acceleration that is exhibited on the foot when it hits the ground.  The way I approached this problem is by using vf^2 = vi^2 + 2ad  and I solved for acceleration.  Knowing that the final velocity is zero I isolated acceleration to get and acceleration of -1024 m/s^2!!! thats huge isn't it.  that is why I'm uncertain about this problem... anyways, multiplying that using the F =ma equation I got a total force of 9216N.  But after inspecting the inital velocity its got to be right because if you convert 6.4 m/s into km/hr you get 23.04 km/hr!!! This person must be falling of a building or something!  So there it is.  Afterall didn't the question start out saying that injury can happen!?