force problem?.. a block with a mass of 4.7 kg is held in equilibrium...?
964943
(4.7)(9.81)cos 20 degree   ... you do the calculator punching..\n\ncheers
510723
Let the weight of the brick be x.\n\nx = 3/4 + (3/4)x\nx - (3/4)x = 3/4\n(1/4)x = 3/4\nx = (3/4)/(1/4)\nx = (3/4)*4\nx = 3 pounds
832683
The energy required to stop the block is it's kinetic energy\n\nE = .5mv^2\n\nThe Energy stored in the spring is\n\nE = .5kd^2\n\nThe two are equivalent, so:\n\n.5mv^2 = .5kd^2\nmv^2 = kd^2\nv^2 = kd^2/m\nv = sqrt(kd^2/m) \nv= sqrt(750(0.012)^2/0.4) = .27 m/sec\n.
545558
Force = mass * acceleration\nForce = 1300 kg * 2.6 m/s2\nForce = 3380 Newtons
706663
The idea is that all forces at the nodes (connecting points) are balanced. Draw the two free body diagrams.\n\nLeft point:\nFx\n0=-T1cos53+T2cos11\nFy\n0=T1sin53-T2sin11-57\nRight point:\nFx\n0=-T2cos11+T3cos42\nFy\n0=T2sin11+T3sin42-W2\n\nBy inspection we see only the first 2 eqns are required. Solve the system:\n0=-T1cos53+T2cos11\n0=T1sin53-T2sin11-57\n\nT1=83.62 N\n{=}8.5 kg
832042
The vertical, horizontal components of all forces should balance.\n       Let the tensions in the ropes be T1 and T2 newtons.\n       T1*cos(90-71) + T2*cos(90-28) = 254.8\n       i.e., T1*sin71 + T2*sin28 = 254.8\n\n       For horizontal componets (which are opp. & equal)\n       T1*cos71 + T2*cos28 = 0\n\nSolve for T1, T2 please.
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"::
833368
find the force of her weight. \nthat should be the answer
852384
I can't help you with the first question, but to figure out the second one you just need to use Hooke's law (I'm assuming the spring to be almost perfectly elastic.) In equation form the law looks like: F=k*x\nWhere F is the force applied, k is the spring constant, and x is the deformation.\nThe force applied will be the objects weight, so 9.8*.5=4.9N, and the deformation is .04m.\nSo 4.9N=k*.04m\nk=4.9N/.04m\nk=122.5N/m
891480
Calculate the k value for the spring.  The stone exerts a force of 8*9.8 = 78.4 N and a compression of 9 cm is\n78.4/.09 = 871 N/m. (Keep it in the MKS system ☺)\n\nWhen the spring is pushed down another 30 cm its total deflection is 39 cm and it's total potential energy is \nkx²/2 = 871(.39)²/2 = 132.5 J\n\nThe stone acquires 132.5 J of gravitational potential energy when released.\n\nSince gravitational potential is mgh, then\n8*9.8*h = 132.5 or h = 1.69meters\n\n\nDoug
703031
Well the first thing to do in a problem like this is extract information from it.\n1.) Weight of Block = 10kg\n2.) Velocity of block(Vi) = 6m/s\n3.) Final velocity of block(Vf) = 0m/s\n4.) Friction acting on block = 20N The friction is always opposite of the direction of the block.\n\nSecond:\nWhat are we calculating? Distance traveled by block! DISTANCE.  We need a formula that will give us distance as an answer! So...we have final velocity, initial velocity, and distance....lets see...how about this one...\n vf^(2) = vi^(2) + 2*a*d \n\nThis equation has Vf, Vi, and D.  \nVf = 0 m/s\nVi = 6 m/s\nD = What we are trying to find.\nThe only thing missing in this equation that we have to find is (a) or acceleration. Hmm.\n\nThere is actually acceleration on  the block.  Friction is acting on the block in the opposite direction; therefore the block is decelerating.  We can use F=ma to find the force of deceleration.\nFriction Force = 20N\nMass = 10kg\nAcceleration = ?\n\nUse the following equation:\nF=ma\n1.) -20N = 10kg * a\n2.) a = -2 m/s (Yes it's negative b/c we are decelerating!)\n\nNow we go back to the original problem!\n vf^(2) = vi^(2) + 2*a*d \n\nThus\n(0^2) = (6^2) + (2)*(-2)*(d)\n0 = 36 - 4d\n-36 = -4d.\nDistance = 9meters! (Make sure to get the units correct!)\nI think I got this right...if not tell me what I did wrong pls\n\n-----------------------------------------------------------------------------------\nThe guy who did it using Energy equations forgot to add one thing....\nmass x (velocity)^2 = Force x distance\nso 10kg * (6m/s)^2 = 20N * x\nso (10 * 36)/20 = 18 meters\n\nThis is another way to do it.  He is using the energy equations to calculate energy in the block. When the block is in motion there is Kinetic Energy in it.  When the block has no motion, there is zero Kinetic Energy in it.  However, the individual that did this messed up the equation as follows:\nKe = (1/2) * Mass * Velocity^2\n\nThus the answer is not 18 but 9 meters!!!!!\nHere is the correction:\n1.) .5*mass x (velocity)^2 = Force x distance\n2.) .5*10kg*(6m/s)^2 = 20N * x distance\n3.) (5 * 36)/20 = distance&#92;\n4.) Distance =9 meters!
840601
(a) Gravity exerts a downward force on the block of 9.8 Newtons/kg * (0.25 kg) = 2.45 Newtons. Work done by gravity = force * distance = 2.45 Newtons * 0.12 m = 0.294 Joules\n\n(b) The spring is a bit tougher since the upward force increases as the spring is compressed. Force = 250 N * (C) where C is the distance the spring has been compressed (in meters). Work is then the integral of Force over distance, so Int(250 * C (dC)) from C = 0 to 0.12 M . The integral is 125 * C^2. Evaluation at 0 gives 0. Evaluation at 0.12 gives 125 (0.12)^2 =  1.8 Joules. So the work done by the spring is -1.8 Joules . Note that the work is negative, because the object is moving against the applied force.\n\n(c) When the block strikes the spring, it will have a kinetic energy of (1/2)mv^2 = (1/2)(0.25kg)v^2 = 0.125v^2.\nThis kinetic energy is absorbed by the spring as it compresses. Note that the work done by gravity is also absorbed by the spring.\nTherefore, we have the following equation:\n0.125v^2 + 0.294 = 1.8\nv^2 = 8(1.8-0.294) = 8(1.506) = 12.048\nSpeed of block just before it hits = SQRT(12.048) = 3.47 meters/sec downward.\n\n(d) I'll leave this as an exercise for you. Just apply the same techniques as above. Note that the block hits the spring with a velocity of 2(3.47) = 6.94 meters per second...
866141
You have to calculate p=massXspeed before the "collision".  The quantity remains constant, so if the mass increases, the speed must decreases by the same ratio for the p to stay the same.  (First, you have to convert the 8N weight to a mass using the 9,8N/kg gravity constant)
824936
a) You have to draw gravity as a straight line down. Then you have to realize that gravity does no work if the block was on a horizontal surface. Work done by gravity is alway M*G*H.  The mass is 4.8 kg and the gravity is 9.81 m/s^2 .  The height will be sin (30) = H/2.  the sin (30 degrees) is 0.5 so .5=H/2.  the height will be .5/2 = .25 m.\n\nsoo you will have 4.8*9.81*0.25\n\nb) the work done by the frictional force will always be the frictional force times the distance.  The frictional force is going to be the normal force times Uk.  The normal force has components to it, so its going to be sin(30)*g*m.  The you need to multiply that by the friction coefficient and then by the distance traveled.  You will have 2*0.436*sin(30)*4.8*9.8\n\nc.  The work done by the normal force is just going to be cos(30) (i think) times 2 ( the distance traveled) times 9.81 (gravity) times 4.8 (the mass)
997326
Use Newton's universal gravitation equation, then sum the forces.\n\nF = G*m1*m2/r^2\n\nF = force (newtons)\nG = gravitational constant (6.6742 x 10^-11)\nm1 = mass of object 1 (kg)\nm2 = mass of object 2 (kg)\nr = distance between objects (meters) ==> don't forget to convert the units\n\nIn this case, you do it twice:\nF1 = G * 390 * 390 / 0.11^2 = 0.84mN to the left\nF2 = G * 390 * 390 / 0.39^2 = 0.067mN to the right \n\nNet Force = 0.84 - 0.067 = 0.772mN to the left
657765
1. the object is pushed up by the weight of its equivalent volume of water, 4-3.2=0.8l\nSo the object's density is equal to its mass divided by its volume:\n4kg/0.8l=4kg/0.0008m3=40,000kg/8m3=5000kg/m3\n\n2. the force decreases with the square of distance, so it will be 3^2=9 times smaller, which is 6N.
211731
If all forces are in equilibrium then the object won't move even if there are several forces acting on it.