find the tension in the supporting cable,if after it was moving downward at 30m/s,it's brought to rest with const.acc.in a distance of 20m.With what force will the feet of the passenger press downward on the elevator floor,when the elevator has an acc.of 30m/s2 upward,if the passenger weighs,a.)16N;b.)640N;c.)64N?
208331
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
377705
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
49670
The struggle here is that you provide too little information to recognize this problem as coherent.\n\nWe may assume that the acceleration of gravity is 9.8 m/s/s or 32 ft/sec/sec.\n\nThe velocity(V) of an object is equal to its initial velocity(v) plus its acceleration(a) times time(t). \nV = v + at or V = v - 32t (in feet per second)\n\nYour object will reach maximum height when the velocity is zero or when the initial velocity is equal to 32 ft/sec/sec * time. With your list of times, you'd just multiply by 32 to solve for the initial velocity, if that is what the problem is actually looking for. There are many other permutations of this data, but hopefully that is a start for you.\n\nGood luck.
372851
a) F = ma \nYou're given F = -6070 N (negative because it's in the opposite direction) and m = 2600 so a = -2.33 ms^-2. Then over 1.76s, the change in velocity is (-2.33)(1.76) = -4.10 m/s. Car velocity is then 34.9 - 4.10 = 30.8 m/s.\n\nb) v^2 = u^2 + 2as\nv = final velocity = 30.8 m/s (from above)\nu = initial velocity = 34.9 m/s\na = acceleration = -2.33 ms^2 (from above)\nRearrange to get s = (v^2 - u^2)/2a = 155.61 m\n\nc) v = u + at\nYou want v = 0. So if the force is constant then acceleration is constant at 2.33ms^-2. \nu = initial velocity = 34.9 m/s\nSo rearrange to solve for time: (v-u)/a = t = 14.98 seconds.\n\nHope that helps. And learn your equations!
395913
F=ma.\n\nSo now you know the available force.\n\nApply it to the mass and use the tangent function to solve.\n\n(Strictly speaking, this is a trick question, since any hill will cause the car to "slow" somewhat.  I get the sense you're missing a parameter, like the minimum speed.)
80915
1) Tension T = m*g = 1240*g\n   work done by tension = T*s = 1240*g*32\n\n2) weight of elevator W = m*g = 1240 *g\n\nthis weight acts opposite to the direction of elevator movment\n\ns0, work = - W * s = - 1240 *g*32
515446
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.
459782
Fnet = Net Force, a = acceleration, m = mass\n\nFnet = a * m\n\n1) Force = 80m/s2 * 0.4kg = 32N\n\nm = Fnet/a\n\n2) m = 330N/5500m/s2 = 0.06Kg\n\na = Fnet/m\n\n3) a = 220N/55Kg = 4m/s2
500194
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.
493517
The frictional force acting against the motion is F\n\n= (mu)x mg x cos a. \n\n= 0.2 x 22 x 9.8 x cos 25 = 39 N. \n\nThe net force acting along the inclined plane\n\n= (mg sin a +39) Since both the forces are acting opposite to the motion.\n\n= 22 x 9.8 x sin 25 + 39\n\n= 130. 12 N. \n\nIf x is the distance moved along the plane, the work done by this net force is Fx.\n\nThis work done is used to remove the kinetic energy of the body = 0.5 m v v.\n\nFx = 0.5 m v v \n\nx = 0.5 x 22 x 3.9 x 3.9 / 130.12 = 1.28m.
548190
Force required to move an object up at constant speed=weight of the object \n\nnow P=WORK/TIME\n       P= FORCE*DISPLACEMENT/TIME=fORCE*VELOCITY\n\nPUTTING VALUES OF P and V\nFORCE=WEIGHT OF ELEVATOR+CONTENT=81000/3=27000N\nOUT OF WHICH\neLEVATOR WEIGHS 18000N\nSO MAXIMUM WEIGHT OF CONTENT IS 27000-18000=9000N\n\nTo get mass divide it with g=10m/s^2 so that you get mass=900 kg
441286
find the force of her weight. \nthat should be the answer
427434
The forces on the person are\n\nSFx = 0 = N - Fc\nSFy = 0 = Fs - m*g\n\nwhere SFx and SFy are the sums of the forces in the x- and y-directions, respectively, N is the normal (to the wall) force, Fc = m*v^2/r is the centripetal force, Fs = u*N is the friction force, m is the mass, and g is the gravitational acceleration.\n\nThese equations say that\n\nN = Fc = m*v^2/r = m*w^2*r\nu*N = u*m*w^2*r = m*g\n\nwhere w = v/r is the angular speed. Solving for the coefficient of static friction gives\n\nu = g/(r*w^2)
553884
For 4 a&b)\nThere would be three arrows.  The first would be a force in the upward direction (i.e. normal force) second would be force due to gravity(downward direction, with equal magnitude o that of the normal force).  The third arrow is an arrow that is in the same direction as the force applied to the large box.  Its magnitude is proportional to the normal force.  It s static friction that is applying that force.   \n\nAs far as part c goes It's been a while for me.  You've got the textbook, look up some formulas.  Besides, I think you'd need to include some masses in order for me to solve those.\n\nThe force diagram for the Large box is very similar to that of the small box, except that it has forces applied to it from the small box. (this is because the large box is supplying the normal force for the small box.)
393809
this question was on this yr's midyear exam, ill give u a clue.\nyou need to seperate the forces acting on the wire into two types \n1) acting horizontally (the force to the left equals the force to the right)\n2)acting vertically (the force to the bottom {or the normal force = 20 *9.8} equals the force to the top)\n\nequate them both and you should get the answer
379130
When the elevator is moving, the tension in the line is equal to:\n\nT = W + F\n\nWhere T is the tension, W is the elevator weight and F is the force associated to the acceleration of the elevator with respect to the building. The forces of W and F are given as:\n\nW = m * g\n\nF = m * a\n\nCombining these equations:\n\nT = m*g + m*a\n\nT = m*(g + a)\n\nSince, T, m, and g are all known, the value of a can be found as:\n\na = (T/m) - g\n\na = (21750 N / 2100 kg) - 9.81 m/s²\n\na = .547 m/s²\n\nHope that helps! :)
122992
k ill try to help u  even though i dont condone cheating. i just know how it is to not understand something and its due in a few hours and it sucks. just remember, nobody will be able to help u on the exam.\n\nso 1. \na. acceleration is change in velocity divided by time. \nsince it started from rest, initial velocity is 0 and then final velocity is given as 25.3 m/s. and the time is given as 0.59s. so to get acceleration   you would do  \na =  Vf-Vi / t\n  =  25.3 - 0 / 0.59\n  = 42.88......\n\nb. the force is equal to mass x acceleration. since you know the mass and acceleration, you just multiply them and you get F = 38421.69... N. \n\nc. you have to find the normal force in this case, which is the force that opposes gravity. like upward force. so in this case, gravity is acting on the driver and force of gravity is equal to mass x 9.8 = 666.4 N. and that means the normal force is 666.4N (since he is not moving up or down the total vertical force = 0)\n\n2. ok well in this case, you know the initial velocity is 0, the distance is 404m and the time is 4.944s. if its constant acceleration, you could assume that the average velocity is just Vi+Vf/2. so if thatz average velocity, then distance = average velocity x time\n     404 = 4.944s  x  (0+Vf)/2\n 404 x 2 / 4.944 = Vf\nVf = 163.43........m/s\n\nthe acceleration would just be Vf-Vi/t. and since you know Vf now  you just do   a = 163.43-0/4.944\na = 33.056....m/s^2\n\n3. ok well acceleration is 9.8m/s^2 (gravity)     distance is 9.7m. initial velocity is 0m/s. so this time, you have to solve 2 equations. the first one is the acceleration equation. a = Vf-Vi / t = Vf / t = 9.8\n9.8 = Vf/t\nVf = 9.8t\nand the other one is the distance = t  x  Vi+Vf/2   average velocity one. \n9.7 = t x Vf / 2\n19.4 = Vf x t\n19.4 = (9.8t) x t\n19.4 = 9.8t^2\nt^2 = 1.979...\nt = 1.406....s\n\nnow that you know the time, you find the Vf by putting it in any of those 2 equations. Vf = 9.8t = 9.8 (1.406)\nVf = 13.788... m/s\n\npart b. so in this case, you use the initial velocity as the one u just found (the velocity of the swimmer at the surface of the water)   so Vi = 13.788m/s\nand Vf = 0    since they come to a stop. and d = 2m\nso you have to find the acceleration. first you need to find the time by using the average velocity equation.\nd = t x Vi+Vf/2\n2m = t x (0+13.788)/2\n4 = 13.788t\nt = 0.29s\nnow that you know the time, you have to find acceleration\na = Vf-Vi/t\n  = 0-13.788 / 0.29\na = -47.5286m/s^2    its negative because its downwards and up is positive in our case. so the force = mass x acceleration and you just got acceleration and her mass is 65kg so you just multiply by 65kg to get F = 3089.36N\n\n\n4. well F = ma and you have the force. when they say it weighs this much newtons, then it means  use gravitational acceleration which is 9.8. so 2350N = m x 9.8m/s^2\nm = 2350/9.8\nm = 239.79kg\n\ni hope you actually understand this stuff. try to do your homework by yourself   dont take advantage of this site. youll just end up screwing yourself over. and if u dont get anythin, let me know and ill try to help