Elastic Energy Problem...NEED HELP!!?
831260
The climber falls through a height h (= 0.75 m) \n\nh = 0.5*g*t^2\n\nSolve for t, the fall time before the rope affects the motion. At the end of that time the climber's speed is\n\nv = g*t\n\nAt the point that the rope runs out of slack, the climber's total energy is\n\nm*g*y + 0.5*m*v^2\n\nWhen the rope runs out of slack it stretches, behaving like a spring, so that when the climber stops falling (briefly) his total energy is now\n\nm*g*(y - x) + 0.5*k*x^2\n\nwhere x is the distance the rope stretched, and (in both the above expressions) y is the height above the ground. Consevration of energy says\n\nm*g*y + 0.5*m*v^2 = m*g*(y-x) + 0.5*k*x^2\n\nSimplifying this leads to\n\n0.5*m*v^2 = -m*g*x + 0.5*k*x^2\n\nand rearranging gives\n\nx^2 - (2*m*g/k)*x - m*v^2/k = 0\n\nThis quadratic equation can be solved for x, the distance through which the rope stretched. Note: be sure to take the physically consistent root when solving the quadratic equation.
905049
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?
834476
No energy is lost as a result of heat or material deformation in an elastic collision.  It simplifies the mathematics by not having to keep track of non-motion energy, but it does not reflect practical results.\n\nIf automobile collisions were totally elastic, body shops would be out of business but funeral homes and hospitals would have a boom.
1116027
Find the potential energy PE=m*g*h\nEquate it to the energy equation\nMass*Height*accln due to gravity=Rise in temp.*Mass*Sp.Heat Cap.\nCancelling mass from both sides and rearranging the equation.\nRise in temp=Height*accln due to gravity/Sp heat capacity\n                    =(200*9.81/5.04*10^2) C
589794
Formulate stability of the system as a function of its energy. Now, one has to show that minimum energy translates into max stability on finding the derivatives.
788598
just taking a shot at it the mechanical energy of the spring will turn to potential energy at the highest point so:\n0.5*k*x^2 = mgh\nx---->compressed distance\nso rearranging h = 0.5k*x^2/(mg) = 5.44m\nhope this is correct and it helps.
986703
In most physics homework problems, energy is conserved, and the following is true:\n\nEo = Ef\n\nHowever, in this problem, energy is lost due to friction.  Therefore...:\n\nEo - Q = Ef\n\nWe know that the child had potential energy at the top, and kinetic energy at the bottom.  Use this information to calculate the energy lost due to friction.  The energy lost due to friction will be denoted "Q" (Q usually stands for heat, which IS thermal energy):\n\nEo - Q = Ef\n\nRearrange the expression, since we're solving for F:\n\nQ = Eo - Ef\n\nNow plug in the formulas for potential and kinetic energy:\n\nQ = mgh - (1/2)mv²\n\nYou don't have to do this, but I factored out the "m" (mass):\n\nQ = m[gh - (1/2)v²]\n\nYou can plug things in now, since the values are all given:\n\nQ = 19.6[(9.80)(3.80) - (1/2)(2.52)²]\n\nQ = 668 J
1037424
evenly distributed energy
858861
1) The gravitational potential energy of water at the top of a waterfall is first converted to kinetic energy (falling) and then converted to both acoustic (sound) and thermal (temperature) energy when it hits the pool of water at the bottom. The temperature of the water in the pool will be slightly higher than the temperature of the water at the top of the waterfall.\n\n2) The thing to remember for this problem is that the energy in a system before some action = the energy in that system after some action:\n\nP0 + K0 + Th0 = P1 + K1 + Th1\nWhere P is gravitational potential energy (mgh), K is kinetic energy (0.5mv^2), and Th is thermal energy (leftover).\n\n22*9.81*4.5 + 0 + 0 = 0 + 0.5*22*2.5^2 + Th1\nSolve for Th1:\nTh1 = 22*9.81*4.5 - 0.5*22*2.5^2 =  902.44 Joules of thermal energy were generated due to friction.\n\n3) Split this problem into two cases, the first 10m and the second 10m. Find the velocity of the crate after the first 10m:\na = F/m = 200/70 = 2.86 m/s^2\nv = sqrt(2*a*d) = sqrt(2*2.86*10) = 7.56 m/s\n\nFor the second 10m, the force of friction on the crate is:\nFf = 0.3*70*9.81 = 206 N\n\nThe net force on the crate is 6N in the direction opposite of movement. The acceleration for the second 10m is:\na = F/m = 6/70 = 0.086 m/s^2\nvf = sqrt(vi^2 + 2*a*d) = sqrt(7.56^2 + 2*-0.086*10) = 7.45 m/s\n\n5) Since watts (W) are measured in J/s and joules (J) are measured in N*m, a watt is the same as N*m/s. Convert lb to N and ft to m to go from horsepower to watts:\n\n550 ft*lb/s * 0.3048 m/ft * 4.45 N/lb = 746 N*m/x = 746 W\n\nPS - You have in your problem description that horsepower is equal to 764 W but that should be 746 W.\n\nI hope this helps!
822266
Conservation of energy:  KE=PE\n1/2*Mb*Vb^2= (Mb+Mw)*g*h\nVb^2=2*(Mb+Mw)*g*h/Mb\nVb^2=2*(0.01+2)*10*0.1/0.01\nVb^2=402\nVb=20.05 m/s
763604
a) The stress in the wire is given by force per unit area = m*g/A\nIf Y is the Young's modulus of the material of the wire, \nY = stress / strain. Therefore \nstrain = stress / Y.\nElongation = strain*original length.\nb) When the object is accelerated upward with an acceleration of a, the force acting in the wire = m*(g+a)\nThe rest of the equations above are valid.\nc) If the elastic limit of the wire is E, the greatest mass that can be accelerated up with an acceleration of a is got from \nE = Mmax*(g+a). Hence\nMmax = E/(g+a)
918781
Where is your freebody diagram? There are other forces acting on the block than just gravity...\n\nGood Luck
676957
E = 1/2 M V^2 where V is the velocity.  Either we need to know the distance of travel of the load or the speed of the vibrating load.\n\nA load  hitting the surface 3 times a second going back and forth 1 ft each way will exert greater energy than a load traveling only 1 inch 3 times a second.\n\nTherefore, if the distance of travel is "d"\nE= 1/2 [M (df)^2]
714752
This problem is a conservation of momentum problem:\n\n(a) In an elastic collision, the formulas for calculating the final velocities are derived from both the conservation of momentum formula and the conservation of kinetic energy formula. Check out my source for more information, but for now just trust me that they are valid:\n\nv1 = (v1(m1 - m2) + 2m2v2)/(m1 + m2)\nv1 = (1*(4 - 7.2) + 0)/(4 + 7.2) = -0.3 m/s\n\nv2 = (v2(m2 - m1) + 2m1v1)/(m1 + m2)\nv2 = (0 + 2*4*1)/(4 + 7.2) = 0.7 m/s\n\n(b) In an inelastic collision, just use the conservation of momentum formula:\n\nm1v1' + m2v2' = (m1+ m2)v\n\n4*1 = (4+7.2)*v\nv = 4/(4+7.2) = 0.36 m/s\n\nI hope this helps!
150677
(The total energy) = (kinetic energy) +(potential energy)\n\nAt the start, most of the energy is potential energy. \n\nAs you slide down, then the potential energy decreases and the kinetic energy increases. \n\nThere is some energy dissipated by friction, that turns into heat. But without friction, the total energy at the bottom of the slide is the same as it was at the top of the slide.
757274
READ THIS CAREFULLY\nAs the block just slides on the table the net force just equals the limiting friction acting on the body\n\nhence,\nF=coeff. of friction * Nor,mal rxn\n\nnormal rxn= weight of body on a plane\n\nhence F= 20*9.8*0.4\n              =78.4N\n\nACC \nit is just begg to slide and the given data is just unsufficient 2 find the acc. of the box
948476
I won't show in details since it's hard to type out the work, but you'll get the rest, here it goes:\n\na)\nWork = mgd  *d=change in height\nW = (10kg) (9.8m/ss) (sin15) (-7.5m) = -190J\n\nb)\nW=(friction coef)(natural force)(distance traveled)\nnatural force = (cos15)(10kg)(9.8m/ss)\nW= (0.4) (cos15) (10kg) (9.8m/ss) (-7.5m) = -280J\n\nc) \nW=Force x distance\nW=(100N)(7.5m)=750J\n\nd)\n(initial KE)+ (puller work)=(gravity work) + (frictional work) + (final KE)\n(final KE) - (initial KE) = 750J - 280J - 190J = 280J\n\ne)\n(final KE) - (initial KE) = 280J\n*solve for final v\nv = squareroot (2 x [280j+ (1/2)(10kg)(1.5squared)] / 10kg) =7.63m/s