total mechanical energy...?
437310
Every mass in motion is said to have mechanical energy.\nHowever a mass is also structure ed with atoms wich are also moving with mechanical energy. The is the energy of the structure. You can represent it a the total mass  time velocity of light squard. Usally in the motion of masses the mass changes are very small. Hence you can consider mass as a constant. The mass time the final velocity squared of the moving mass is the mecahnical energy of maas in motion. assuming there is no friction. if there is friction you must substract the energy loss due to friction.
865501
This is letting spring energy = gravitational potential energy.\n\nkx²/2 = mgh→ k = 2mgh/x² = 2*1*9.8*.6/.02² = 29400 N/m
949133
OMG!\ni have the same question!\nmine is:\nA piece of an unknown metal has a mass of 20.0 g and at 100.0 C is placed in a styrofoam cup calorimeter containing 50.7 g of water at 22 C. If the final temp is 27.2 C, calculate the specific heat capacity of the metal. Assume that all the energy lost by the metal is gained by the water.\n\nok, haha fine, not the SAME question, but close enough. But, i'm uber confused. ><
825498
Kinetic Energy = 1/2mv^2; Potential Energy = mgh\n\n1/2mv^2 = mgh or 1/2v^2 = gh\n\nv^2 = 2gh = 2 * 9.81m/s^2 * 0.49 m\n\nv = 3.1 m/s
43971
Da_vinci's answer is partly correct but it confuses force, power and energy. Energy is the integral of power over time, or more simply, if power is constant, energy is power * time. Kinetic energy can also be expressed as .5*M*V^2, M being mass and V being velocity.\nDa_vinci's example illustrates that since velocity is measured with respect to some reference, so is kinetic energy. Without a reference an accelerated object receives a velocity change but the kinetic energy change cannot be defined. Similarly, without a reference, decelerating the object requires application of a force but does not recover any energy.
83134
you really should have given numbers \n\nkinetic energy will be 1/2mv^2\n\nIts linear speed can be determined by distance over time  (d/t)\n\nthe next 3 answers need me to know what data you have in order calculate their values
791539
Wow. Your classmate just asked this question. But here's your answer anyways...\n\nThere are a few things to keep in mind here: all of the energy in the initial system is rotational, and the initial energy is equal to the energy lost to friction. Also note that the energy is lost to friction at a constant rate.\n\n(a) The initial energy of the system is:\n\nE = (1/2) * I * w²\n\nWhere I is the moment of inertia and w is the angular velocity. The power transferred from the system due to friction is equal to the time derivative of the energy loss:\n\nP = dE/dt = I * w * a\n\nwhere a is the angular decceleration.\n\nSince the power is transferred at a constant rate and removes all of the energy from the system, we can take this as:\n\nP = -E/t (negative sign denotes power loss)\n\nCombining the power equations yields:\n\nI * w * a = -E/t\n\nSolving for a:\n\na = -E / (I * w * t)\n\nFinally, substitute the expression for E to obtain:\n\na = -I * w² / (I * w * t)\n\na = -w/(t)\n\nSubstituting the values yields:\n\na = -(150 rev/min * 2*pi rad/rev * 1/60 min/sec)/(2.2 hr * 3600 sec/hr)\n\na = -(15.71 rad/s)/(7920 s)\n\na = -.001983 rad/s²\n\n(b) The rotational behavior of the fly wheel can be modeled as simply:\n\nR(t) = -1/2*a*t² + w*t + Ri\n\nwhere R(t) is measured in radians and Ri is the initial angular displacement. Assuming that Ri = 0 (we'll measure the number of rotations from its initial position), then the flywheel will revolve:\n\nR(t) = -1/2*a*t² + w*t (radians)\n\nR(t) = (-1/2*a*t² + w*t)/(2 * pi) (revolutions)\n\nSubtituting the values:\n\nR(2.2 hr) = R(7920 s) = (-.5*.001983*(7920 s)² + (15.71 rad/s)(7920 s))/(2*pi)\n\nR = 9900 revolutions\n\n(c) The tangential component of the linear acceleration of a point on a rotating body is given as:\n\nA-tangent = a * r\n\nwhere A = linear acceleration. The angular acceleration is constant, so for all w (including w=75 rev/min), the linear acceleration will also be constant. Thus:\n\nA-tangent = (-.001983 rad/s²)(.5 m)\n\nA-tangent = -9.915 x 10^(-4) m/s²\n\nA-tangent = -0.9915 mm/s²\n\n(d) The magnitude of the net linear acceleration on the particle is composed of a tangential and a normal component. The equation is of the form:\n\nAnet = [ (A-tangent)² + (A-normal)² ]^(1/2)\n\nOr:\n\nAnet = [ (a*r)² + (v-tangent²/r)² ]^(1/2)\n\nNote that the tangential velocity is given by:\n\nv-tangent = w * r\n\nThus, the magnitude of net linear acceleration is:\n\nAnet = [ (a * r)² + (w² * r)² ]^(1/2)\n\nSubstituting the known values yields (note that for this case, the angular velocity of 75 rev/min is equal to 7.854 rad/s):\n\nAnet = [ (-.001983 * .5)² + (7.85² * .5)² ]^(1/2)\n\nAnet = 30.81 m/s²\n\nFINAL ANSWERS:\n\n(a) a = -.001983 rad/s²\n(b) R = 9900 revolutions\n(c) A-tangent = -0.9915 mm/s²\n(d) Anet = 30.81 m/s²
746965
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
816057
again ... conservation of (mechanical only here) energy\n\npot energy = kin energy\n\nmgh = 1/2 m V^2\n\nso V = sqrt(2gh) -- see mass irrelevant\n\nV = sqrt(2*9.8*8.53) = 12.9 m/s\n\ng=9.8m/s^2 is the acceleration due to gravity\n\ng=10m/s^2 is not a bad approximation
772410
electrical energy-------> heat + light\ndoes that help?
641698
The total amount of energy in a closed system remains constant.\nThis, however, was found to be untrue. It was later substituted for the conservation of Mass-Energy.\nThis was also disproven. In fact, if a tiny bit of energy is "borrowed" from nothing for a really short amount of time, it can create mass according to the equation:\nE*t = h/(2*pi) = 1.06*10^-34 Js\nOf course, for our pruposes, the conservation of energy works.
859502
What you are likely looking is W=F*D \n\nWork = A Force * A Distance\n\nCheck out wikipedia for additional help:\nhttp://en.wikipedia.org/wiki/Mechanical_work
916732
Is relative to the amount of energy and object contains.
767789
decrease potential energy.....through physical factors like friction....
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
573522
kinetic energy is 1/2 mv² so it depends on mass and velocity
67988
The first answer is correct.  Energy.  Energy is defined as force applied through a distance.  This specific type of energy is known as kinetic energy (energy of motion).  Energy is measured in joules, or J.  E = F x D, so if you applied 50 Newtons and the object moved 1 metre, you would have expended 50 joules.