An object with a mass of 100 Kg is dragged up an incline by applying a force F. The total distance moved along
961494
This is an energy balance situation.  The work done in dragging the object must equal the energy change.  In this case the energy is potential energy, given by m*g*∆h, where ∆h is the change in vertical distance.  The work (energy) required to drag the block is force times distance along the incline.  Equate these two quantities and solve for F.
1093836
Centrifugal force and gravity.
621321
If the density is 'd' kg/m^3 then the weight is '9.81d' Newton. [remember weight is force]
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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.
898267
Given: mass, m, of crate=14kg\n   therefore, weight of crate=14g\n   initial speed of crate=1.56m/s\n   applied force along the incline=150N \n   distance traversed by crate=7.99m\n   mu=0.28\n   acceleration of gravity,g=9.8m/s^2\n   angle of incline=25 degree\n\na) Initial kinetic energy, KE=1/2mv^2\n                =1/2*14*1.56^2=17 Joules\n\nforce of friction=mu*R, \nwhere R=14*9.8cos25=124.3N\n\nNet force,F=150-124.3\n                  =25.7N\n\nF=ma\n25.7=14a\na=25.7/14\n  =1.84m/s^2\n\nv^2-u^2=2as\nv^2-1.56^2=2*1.84*7.99\nv^2=29.4+2.43\n     =31.8\nv=5.64m/s\n\nFinal KE=1/2*14*5.64^2\n              =222.6 Joules\n\nChange in KE=222.6-17\n                      =205.6 Joules\n\nb) 5.64m/s as computed in a) above.
903133
From what I learned in physics class, the force you are speaking of is measured in Newtons.
979740
During the first 10 m the force increases in a linear fashion. Because it's linear, you can average it for the first 10m: 5 N. This easy way will give the same velocity at the end of the 10 m as if you did it by area under the curve. According to Newton's 2nd law, this gives the object acceleration a1:\nF = m*a1 where F = 5 N and m = 6.90 kg.\n\nUse a1 in the equation V^2 = Vo^2 + 2*a1*10 m where Vo = 0. Solve for V.\n\nFor the second 10 m section, the force is steady at 10 N. Again use Newton's 2nd law to get a2:\nF = m*a2 where F = 10 N and m = 6.90 kg.\n\nUse a2 in the equation V^2 = Vo^2 + 2*a2*10 m where Vo = V from the last part.  Solve for V.
816694
hi \nbasic formula to be used here is f=ma\nhere we have 2 scenarios elevator going up(against gravity of earth denoted by 'g' which is constant eual to 9.8 or lets say 10m/s ^2.\nnow when elevator is going up means it has to defy gravity and go up which means there has to be more force to accomplish\nhence we use f=70 x (g+a)\nwhich is f= 70 x(10+3)=910N\nwhile the elevator is moving down it doesnt need to defy gravity and hence the force required to accomplish this will be much less\nhence we use f=70x (g-a)\nwhich is f=70x(10-3)=490N\n\nis the elevator is standing still then \nf=m x a\nwhich is f= 70 x 3 = 210n
12996
I can't draw a free-body-diagram here, so it's going to be somewhat more difficult to explain (a picture is worth 1000 words and all)...\n\nSince the block is moving at a constant velocity, we know that all the Forces are balanced.  this means that the Force of friction is equal to the force of gravity pulling the block down the incline.\n\nI'll write this as \nFf = uN  \n\nSince we know the angle of the slope, 25 degrees, we can break the Force of gravity into 2 components,  one that points down the incline (i'll refer to it as Fx) and one points the opposite direction (I'll refer to is as Fy) of the Normal vector N.\n\nIf you draw that triangle you notice the the hypotenuse has magnitude(length) of mg (mass times the gravitational constant.\n\nnow, lets use the sin & cos relationships to determine the length of the 2 components: Fx & Fy\n\nsin = opposite / hypotenuse\ncos= adjacent / hypotenuse.\n\nIf you look at the diagram, the opposite is Fx so\nsin 25 = Fx / mg \nFx = sin25 * mg \n\nand the adjacent is Fy so\ncos 25 = Fy / mg\nFy = cos25 * mg  \n\nagain since all force are in balance, \nFy = N \n\nand Fx = uN \nsin25 * mg = u * Fy \nsin25 * mg = u * cos25 * mg \nwhich reduces to\n\nsin25 / cos25 = u \nand \nfinally\nu = tan25 = ~.466
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Net force on object = Force applied - Friction force\n\nSince velocity is constant, the acceleration of the object is zero\n\nSo, Net force = mass * acceleration = zero\n\n0 = Force applied - Friction force\nor\nForce applied = friction force \n= coefficient of friction * Normal force\n=.250 * 11.9 kg * 9.8 m/s^2\n=29.2 kg-m/s^2\n\nBox is decelerating at .250*9.8m/s^2\n\nand\n\ndistance traveled = (final velocity^2 - initial velocity^2)/2a\n=(0 - 3.2 m/s)  /  2*(-.250*9.8m/s^2)\n=.65 m
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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
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Getting started, have to overcome static friction.  Therefore, force applied = static friction force just before movement\n\nF = umg\nF = 75N\nm = 20kg\ng = gravity\nfind u, plug and play\n\nsame problem for sliding friction
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a=V^2/r, so the larger the radius the larger the acceleration and since F=ma and assuming mass stays constant the force must increase also
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If i understand your scenario, the force is pushing downward on the block.  The horizontal component of the force is 15 N times cosine of 40 degrees.  The vertical component of the force is 15 N times sine of 40 degrees.  However, there is also the force of gravity, which is (3.5 kg * 9.8 m/s2)  (units are N, like the other forces).  So if you add the force of gravity to the vertical component, you have the total downward force.  Multiply this by the coeficient of friction, and you have the force of friction.  \n\nThe force of friction is directed against the direction of motion.  Subtract the frictional force from the horizontal component, and you have the total horizontal force.  Divide the horizontal force by the mass and you should get the acceleration.\n\nfriction force = [(15*sine 40)+(3.5*9.8)]*.25\nacceleration = [(15*cosine 40)-friction force]/3.5
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Work done = force x distance moved by object i thinkin a straight line ,so ithink some conversion maybe needed.
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We did something like this in my physics lab with an Atwood machine.  It was a pully with weights hanging from it.  I can't exactly remember how it worked, but maybe this will inspire you.  We also did something with an air track and these two masses colliding and we'd see which one would go in which diretcion.  You need a nearly frictionless surface, like an air hockey table or ice slab.
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1) The angle of the rope needs to be considered to calculate horizontal and vertical components of the clerk's force. \nFh = 188*cos(25) = 170 N This is what pulls the box.\nFv = 188*sin(25) = 79.5 N This actually lifts the box slightly.\n\nThe symbol for coefficient of kinetic friction is µk. A lower number makes it easier to move the box - if you spilled some olive oil on the floor µk would perhaps go to 0.1. N is normal force. If this was on a slope, normal would be different from weight, but the floor here is level so N = W? \n\nActually no. Because the rope pulls up on the box slightly, the normal force pressing the box to the floor is less than the box's full weight. The box's weight = m*g = 36.9 kg*9.8 m/s^2 = 362 N. The vertical component of the tension in the rope relieves 79.5 N of that so N = 362 - 79.5 N = 282 N. \n\nSo Ff = .450*282 N = 127 N is the force that friction fights with. The horizontal portion of the clerk's pull is 170 N. So there is 43 N left over to accelerate the box. Newton's 2nd law says\nF = m*a\n43 N = 36.9 kg*a\nso a = 1.17 m/s^2\n\n2) Here the force is directed down at an angle so the vertical component of the force will add to the normal force used in calculating the friction.\nFh = 435 N*cos(35) = 356 N\nFv = 435 N*sin(35) = 250 N\n\nN = the weight of the box + Fv = 259 N + 250 N = 509 N\n\na) Ff = µk*N = .57* 509 N = 290 N This is not enough to hold the box stationary so 356 - 290 N = 66 N are available to accelerate the box. Newton's 2nd law again:\nBut first we need the box's mass. W = m*g\nm = W/g = 259/9.8 N = 26.4 kg. Now\nF = m*a\n66 N = 26.4 kg*a\na = 66/26.4 m/s^2 = 2.5 m/s^2\n\nThis is getting complicated. To move 9 m will take t seconds:\nx = (1/2)*a*t^2 Solve for t.\n\nb) Ff = µk*N = .75* 509 N = 381 N This is more than the horizontal portion of the clerk's push, so the box will resist. The answer is 0.