How to construct a speed ramp which converts gravitational potential energy of a ball into kinetic energy?
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You stick books under one end of a plank and roll the ball down it.
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To solve this problem, you need to use conservation of energy.  At the beginning, the block is at rest at the top of this ramp.  At the bottom of the ramp, the block still has some kinetic energy, but some of it has dissipated due to friction.  Thus, the equation that governs this problem is:\n\nInitial Potential Energy = Dissipated Energy due to Friction - Final Kinetic Energy\n\nSince Potential Energy is defined only in the vertical direction, we must use the vertical components of the energy.\n\nWe can find the height of the ramp using trigonometry:\n\nsin(theta) = height/incline -----------> height = incline*sin(theta)\n\nheight = 3.03*sin(28.8) = 1.46 m\nPotential Energy = mgh = (2.51)(9.81)(1.46) = 35.95 J\nFrictional Energy (vertical) = u*m*g*sin(theta)*h\nKinetic Energy (vertical) = 0.5*m*[v*sin(theta)]^2 = (0.5)(2.51)(1.77sin(28.8))^2\n\nKinetic Energy = 0.91 J\n\nPlugging these values into the equation, you get:\n\n35.95 = u*m*g*sin(theta)*h + 0.91\n\nWe have all the values, now solve for u\n\n---------------\n\nHope this helps
887598
Well a very very simplistic demonstration could be shown on a pool table.\n\nIf you shoot the cue ball at a stationary ball (let's call it the 8 ball) across the table, the cue ball has Kinetic Energy, where as, the 8 ball has only potential energy.  \n\nWhen the cue ball collides with the 8 ball, the kinetic energy will transfer from the cue ball to the 8 ball, sending the 8 ball in motion.  Now that the 8 ball is in motion it has Kinetic energy which is a change of form.  \n\nThe Law of Conservation of energy says that energy cannot be created.  So what does that mean?  It means that if the 8 ball is now in motion, it took energy from the cue ball, which means that the cue ball has now lost energy, and will be traveling slower, and if they hit at the right spot (since they are approximately equal in mass) the cue ball can even come to a complete stop upon collision.\n\nHope I didn't lose you!\n\n~~Tyler\n\n\nAlso, an easier example would be a monkey falling out of a tree.\n\nIf the monkey is sitting down eating a banana on the highest branch, he has Potential energy.  If he reaches for another banana and it's just out of his reach and he slips off of the branch, his potential energy is now being converted into kinetic energy.\n\nHopefully theres a pile of leaves on the ground.
705009
The total energy of the hoop as it enters the ramp is the sum of its forward kinetic energy and its rotational energy.  If it is rolling at w rad/sec, it is moving forward at a speed of 2*π*R*w.  The total kinetic energy is then .5*M*[2*π*R*w]^2 + .5*I*w^2   I is the moment of inertia of the hoop, M*R^2.\n\nThe total kinetic energy is converted to potential energy when the hoop stops.  Its potential energy is M*g*h, where h is the height of the hoop above the horizontal portion.  If the hoop travels a distance L up the ramp, it climbs an amount = L*sinø.  Now equating the potential energy at the top to kinetic energy at the bottom you get\n\n.5*M*[2*π*R*w]^2 + .5*M*R^2*w^2 = M*g*L*sinø\n\nYou have all the data to solve for L
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1. Usually you have to solve these problems in terms of energy, where initial energy = final energy + friction (lost) energy. The initial and final energies are each the sum of the respective initial and final potential + kinetic energies. (Sometimes the problem doesn't explicitly give you an initial potential energy; it may just tell you that an object slides x meters downhill.) If you know the initial and/or final momentum and either the mass or the velocity, you can find the missing variable (velocity or mass) and then transform the momentum into kinetic energy. \n2. The coefficient of friction 'mu' of course affects the friction force. Sometimes you are asked to find mu so you have to work backwards. You may know the lost energy and the length over which the object encountered friction. Then you solve for friction force. Then, knowing the normal force as well (force perpendicular to the friction surface), you can find mu.\n3. Inclines with friction are a big favorite of the problem writers. Just remember that the gravitational force normal to the incline is g*m*cos(theta) and the gravitational force along the incline is g*m*sin(theta). If a ramp is tilted just enough to allow the object to slide, g*m*sin(theta) = friction force = mu*g*m*cos(theta), so mu = sin(theta)/cos(theta) = tan(theta). Or if you know mu and want to find theta, it's arctan(mu).\n4. Another fave is centripetal force on a horizontal turntable with a friction surface. We know the centripetal force is w^2*r*m (w being angular rate in rad/sec). And we know the normal force is g*m. So, if we rev up the turntable until the object slides off, or we move the object out from the center until it slides off, we know that, at the instant it lets go, w^2*r=g*mu, so we can find w or r. Easy, right?\nI hope this helps; if you have followup questions I'll be watching this question until it closes.
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Kinetic Energy = 0.5 * mass * velocity^2\n\nKinetic Energy = 0.5 * 4000kg * (3m/s)^2\n\nKinetic Energy = 18,000 Joules
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This problem employs the simple concept of conservation of energy (athough there are a number of ways to solve it).\n\nBasically, the person in the wheel chair has an initial kinetic energy at the bottom of thr ramp,\nKE = 1/2 mv^2\nAt the top of the ramp, some of that kinetic energy has been transfered into Gravitational Potential Energy,\nPE = mgh\n\nSince the mass of the person is constant, that drops out of the equation.\n\nSolve for the change in PE and subtract this from the intial KE.  Using the new KE, solve for the final velocity.
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For moving objects, calculate kinetic energy using the formula:\nKE = (mass x velocity2)/2 or 1/2 mv2\n\nhttp://www.gmhsscience.com/problems/kineticpotential.htm
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Potential Energy:  A car is at the top of a steep hill, in neutral, wheels straight,  with the parking brake on.  Release the brake.  The car will roll downhill, converting potential energy into kinetic energy.\n\nKinetic Energy:  The car continues rolling onto a flat stretch of road, which happens to be covered with grapes.  As it rolls over the grapes the car squishes them and slows down.  It's kinetic energy is being transformed into grape juice and heat.
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For a back row spike I found 26.7-27.6m/s (mean).  Sorry, but that was all I could find!  I was going to do a study on it - but then I switched out of biomechanics into exercise physiology.  You could do it by calculating time elapsed between contacting the ball and the ball reaching the floor.  You would have to know the height at which the ball was contacted and how far it traveled horizontally from the point of contact.  They you find the hypotenuse using Pythagorean theorem and divide this number by the time to get an estimated velocity.
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Vertical components\n\nv = u + at\n0 = u + (-32)(2.1s)\nu = 67.2ft/sec\n\nMaximum height the ball reached\n\nv^2 = u^2 +2gs\n0 = (67.2)^2 + 2(-32)(s)\n64s = 4515.84\ns = 70.56\n\ndownward velocity 0.8 seconds after the ball reached the max.height.\n\nv = u +at\nv = o +(32)(0.8)\nv = 25.6 ft/sec\n\ndistance from the maximum height\n\nv^2 = u^2 +2gs\n(25.6)^2 = 0 + 2(32)s\ns = 10.24 ft\n\nHeight of the fence = 70.56 ft - 10.24 ft\n\n\nHeight of the fence is 60.32 ft...\n\nthere ya go...
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I would use conservation of energy.\nx is the distance traveled on the ramp\nthe forces at work are:\nGravity.  The potential energy will be\nsin(5)*m*g*x\n\nFriction:\n.48*cos(5)*m*g*x\n\nSince there are no brakes all of the kinetic energy will be lost to friction and the potential energy gain:\n\n.5*m*v^2=\n.48*cos(5)*m*g*x+sin(5)*m*g*x\nnote that m divides out\nI get 110 m\n\nj
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Kinetic energy.\n\nGreek kntikos, from kntos, moving, from knein, to move.
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For a sphere of constant density having mass M and radius R, the moment of inertia is\nI = (2/5) M R^2\nThe kinetic energy of a travelling and spinning sphere of uniform density is\nK = (1/2) I w^2 + (1/2) M v^2\nK = (1/2)(2/5) M R^2 w^2 + (1/2) M v^2\nK = (1/5) M R^2 w^2 + (1/2) M v^2\nWhere w is the angular speed.  And where v is the speed of the sphere's center of mass with respect to the surface on which the sphere is moving, which is equal to the tangential speed of the sphere's rotational "equator," which is in contact with the surface on which the sphere is moving, with respect to the sphere's center of mass.\nw = v / R\nK = (1/5) M R^2 w^2 + (1/2) M v^2\nK = (1/5) M v^2 + (1/2) M v^2\nK = 0.7 M v^2\nK = 0.7 M R^2 w^2\nFor your sphere, M=0.5 kg and R=0.06 m, so...\nK = (0.00126 kg m^2) w^2\nOn a slope 30 degrees from horizontal, the acceleration from gravity would be\ng = (9.8 m/s^2) cos(30) = 4.9 m/s^2\nThe work done by a ball rolling one meter along the slope is\nE = M g (1 meter) = 2.45 Joules\nK = E\nThe angular speed of the sphere after rolling 1 meter is\nw = { K / (0.7 M R^2) }^(1/2)\nw = { 2.45 Joules / (0.00126 kg m^2) }^(1/2)\nw = 44.096 s^-1\nv = wR = 2.6458 m/s\nThe average acceleration of the ball is found from the regular kinematic formula.\nVf^2 - V0^2 = 2 a dx\nWhere dx is the one meter measured along the slope, V0=0, Vf=v.  The average acceleration, therefore, is\na = v^2/(2 meters) = 3.5 m/s^2\nSo the speed of a sphere of constant density having mass 0.5 kg and radius 0.06 meters, rolling freely down a 30 degree frictionless slope, is\nv = (3.5 m/s^2) t
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Basically you have the right idea. Use momentum (mass x velocity) for the motion and the sudden change in speed and/or direction (acceleration) and you get force x time = kinetic energy trying to rip you apart or squash you like a bug.
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You should only ask one question at a time.\n\nI will answer the first:\n\nConsider the forces on the box:\n\nGravity vertically downward\nfriction pulling backwards\nThe force of gravity parallel to the ramp \nminus the force of friction\n=m*a=m*1.22\nso\nm*sin(34)*g-f=m*1.22\nfriction can be expressed as\nf=m*u*cos(34)*g\nnote that mass divides out\nsin(34)*g-u*cos(34)*g=1.22\n\nu=.52\n\nj
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First assume that we neglecting any losses from air resistance or internal friction.\n\nWith that simplification, imagine two states:\n1) you are on the bridge 100m above the water and at rest\n2) you are suspended by the spring at its point of maximum extension, and therefore with zero velocity\n\nSince we assumed that there were no losses, the energy of both of these states is equal. Where is that energy located?\n -->1: your gravitational potential energy (no kinetic energy, no stored elastic energy)\n--> 2: your remaining gravitational potential energy, and the stored elastic energy in the spring (by definition, no kinetic energy)\n\nSo what has happened is that the spring has stored elastic potential energy that is equal to your loss of gravitiational potential energy. That's the only way the energies of the two states are the same.\n\nNow you can set up some equations. Let's say that you start at height H0 and end up at height H1.\n\nchange in PEg= m * g * (H1-H0)\nchange in PEk = 1/2 * k * [(H1-H0)-30m]^2\nwhere we have accounted for the 30m long unstretched length of the spring.\n\nNow by observing that -PEg = PEk, and plugging in the given values for m and k and the known value for g, you can readily solve for the change in height (H1-H0) (it should be < 0) It's a quadratic, so you expect to find one answer that is not physically realizable - reject that one, and select the other one.\n\nDoes that help?