Lee pushes a 21 kg mass 10 m across a floor with a horizontal force of 70 N. Calculate the amount of work Lee?
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Part 1: Mike\nWork=Force x Distance(moved) x cos (theta) (angle between work and distance)\nF=215\nD=64.7\nTheta=35 degrees\n\nW=215*64.7*cos35\nW=11,394.815 Joules\n\nPart 2: Chris\nF=8N\nD=3.9\nTheta=90 degrees\nW=8*3.9*cos90\nW=0\n\nThats all you get...good luck
932992
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.
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(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...
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OK, the barrel´s mass is 40Kg (notice: its mass is 40 Kg, not its weight). \n\na) So, your formula is Weight = mass(m)*gravity´s acceleration(g)= 40Kg*10m/s^2 = 400 KG*m/s^2 (=400 Newton)\n\nb) Work = force * distance\nSince work is calculated only for the vertical, and the force equals the weight, you have:\nWork = 400N*5m = 2000 Joule\n\nc) Eff = 250N*5m/2000J*100 = 0.625*100 = 62.5%\n(You got 12.5% because you forgot to multiply 250N for 5 meters. Remember that efficiency is a relation of works, and work is force * DISTANCE).\n;o)
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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.
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Work = F * d\nWork = Force of the spring * distance mass moves\n\nWork refers to an activity involving a force and movement in the directon of the force. A force of 20 newtons pushing an object 5 meters in the direction of the force does 100 joules of work.\n\nThe max KE is the work put into moving the object.
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Fnet=M*A \n\nYou know that the only forces are friction and the applied force of the orderly because friction and the normel force cancel out.\n\nso:\nFo-Ff=M*A\n70N - xN=80Kg * .4m/s^2\n70N - xN=32Kg*M/s^2 =32N\n-xN=32N-70N\nx=38N\n\nThe force of friction is 38.0N in the direction opposite of motion.
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Energy Unit = Joule = 1 kg x m^2/s^2 = 1 N x m\n\nHow can work transfer into keinetic energy?\nThink in a pendulum in which these two are constantly tranfered into one another.\n\nWork = Force x Distance \n\nwork = 10N x 2m = 20 N x m\n\nKinetic energy = Mass x Velocity^2 / 2\n\nkinetic energy = 2kg x (4m/s)^2 / 2 = 16 N x m
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find the force of her weight. \nthat should be the answer
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You don't need to find the mass if use conservation of energy to solve the problem.\n\nThe kinetic energy of the can initially is:\n1/2*m*v^2\n=.5*m*2^2\n=2*m\nand when it comes to rest it is zero\nThe energy was lost to the work done by friction.\n\nthe force of friction times displacement is the work:\n\nu*g*m*d\n=.2*9.8*m*d\n=19.6*m*d\nnow set the energy of the system equal to zero:\n2*m-19.6*m*d=0\ndivide out the mass and solve for d:\nd=19.6/2\n=9.8m\n\nj
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using conservation of energy the kinetic energy at the bottom is \n1/*m*v^2\nThe frictional force times the distance is the work done by friction\nand the displacement in the vertical direction is the potential energy that gets converted to kinetic and friction\n\n1/2*m*v^2+f*L=m*g*h\nwhere h /L= sin(15.9)\n\nPlug in the numbers and solve using algebra\n(.5*6500*3.8*3.8)+4300*L=(6500*9.8*sin(15.9))*L\n46930+4300L=17451*L\n46930/(17451-4300)=L\n=3.6m\n\nj
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you need to identify that there're two parts of the force, one acting horizontally, the other acting vertically.\nSo:\n1) the vertical part is:\n70sin30\n2) tje horizontal part is:\n70cos30
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Force = mass * acceleration\nForce = 1300 kg * 2.6 m/s2\nForce = 3380 Newtons
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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
738706
Both questions use the formula F=ma (Force = Mass * Acceleration):\n\n1) F = ma\n1310 = 214*a\na = 1310/214 = 6.1215 m/s^2\n\n2) F = ma\n3.6 = m*9.8\nm = 3.6/9.8 = 0.367 kg
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Frictional force, F = u*m*g\nHere, F = 300 N, u = ?, m = 80 kg, g = 10 m/s/s (assume)\nTherefore, 300 = u*80*10\n=> u = 3/8
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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