What happens to the kinetic energy of a falling book when the book hits the floor?
91362
It's tranferred to heat/friction from the impact
892916
I think the key here is that the objects will be rolling down the plane; thus the gravitational potential energy they have is not only converted into translational kinetic energy, but also into rotational kinetic energy.\n\nAnyway, I would approach this from energy considerations.  Start by calculating the amount of potential energy the objects will lose by falling a distance of 3 m.  (For both, this will simply be PE = m g h).  Equate this to the total amount of kinetic energy the objects have at the bottom of the ramp (translational plus rotational, or 0.5 m v^2 + 0.5 I w^2 where I is the moment of inertia, and w is the angular velocity.)\n\nSo, you need to know the moment of inertia for a disk and a hoop.  The hoop I know off the top of my head, I(hoop) = m R^2.  The disk, I looked up to find I(disk)=0.25 m R^2.\n\nFinally, recognize that as they do not slip, you can relate the angular velocity w, to the translational velocity v:  v=2 Pi R w.\nOnce you do this, you find that the kinetic energy depends only on the velocity v, so we have one equation with one unknown.  Just solve for v with the two different values of I, and I think you ought to find that the disk will be moving faster than the hoop.\n\nSorry for being long winded.... but I hope it helps.
735384
Using the law of conservation of energy determin\n\na) the vertical component of vector V(initial)\n\nIt went up 20 meters, mgh = 1/2mV^2, solve for velocity\n\nsqrt[2(9.8m/sec^2*20meters)] = v = 19.8m/sec\n\nb) the work done by the gravitational force on the particle during its motion from P to the ground.\n\nMaybe this is a trick question, I will take it at face value, when at P it is 60meters above the ground, the work of the gravity bringing it down from P is just mgh = 0.5kg*9.8m/s^2*20m=98J\n\nc) the horizontal and vertical components of the velocity vector when the particle reaches the ground:\n\nHorizontal velocity is constant at 30m/sec, nothing to calculate.\n\nVerticle velocity, keep in mind that the object reached its maximum height at 80 meters high (ie from a verticle perspective is is at zero velocity at that point, and begins to fall, the terminal vertical velocity comes from the potential energy of \nmgh = 0.5kg*9.8m/s^2*80m = 1/2 mV^2, solve for v\n\nvertical velocity is 39.6 m/sec downward\n\nplease check for calculation errors.
686704
The increase of the gravitational energy (m1+m2)gh is the loss of kinetic enery m(v^2)/2.\nThe bullet gets the same speed as the bob.\nUse conservation of momentum.
470687
At the top of the halfpipe, the ball is just sitting there with potential energy stored up (PE = mgh, where m = mass, g = acceleartion due to gravity, and h = height).  As the ball rolls downward, its PE lessens and its kinetic energy increases (KE = 1/2(mv^2), where v = velocity of the ball).  At the bottom of the halfpipe, all PE is gone and KE is at maximum.  As the ball goes up the other side, the KE decreases and the PE increases.  In a perfectly frictionless halfpipe, the ball would return to the same height as it was before it started rolling, and the KE would go to 0 and the PE would be at its maximum.\n\nIf you just dropped the ball (no halfpipe - just a freefall), the KE just before impact would be the same as the PE before dropping.  The two are inversely proportional - as one goes down, the other goes up.\n\nNow if you knew the exact shape of the halfpipe, you could start plugging in numbers to find the PE and KE of the ball at various points.
896193
This is an impulse/momentum problem.  Impulse is ∫f*dt[0 - t1] it must equal the change in momentum.  Impulse is given by Favg*t, in your case 0.012*16000 N*sec  Therefore the momentum of the student on impact was 0.012*16000 = m*v.  m is the mass of the student, 63kg.  Therefore his velocity on impact was v =0.012*16000/63.  He fell under acceleration g = 9.8m/sec^2.  Velocity after falling time t is g*t.  Distance traveled under uniform acceleration is h=.5*g*t^.    Use the velocity to get t, then compute h.
208645
All objects accelerate downward at a rate of 9.8 m/s/s.  If the object falls 2.6 m, at that rate, it would it the ground in approximately .53 seconds at a speed of 5.2m/s. To calculate it falls approx. 4.9m the first second then continues to speed up at a rate of 9.8m/s/s after that.  Since 4.9m is farther than the distance fallen, divide 2.6 by 4.9 and get the time taken to fall.  Take that answer and multiply that by the constant rate of acceleration to get 5.2m/s
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
1032909
Total energy at the start is equal to the gravitonal potential energy plus the initial kinetic energy. That equals 320000j + 14400j. At the bottom of the hill, the kinetic energy is 160000j and there is no more potential energy. The amount of energy lost to friction, in the form of heat, is 174400 joules.
209055
Potential Energy: Energy due to position.\nKinetic Energy: Energy due to motion.\n\n======================================================\nAs the ball moves up it gains potential energy and loses kinetic energy. It the maximum height, it has all potential energy and no kinetic energy. As it moves down, it loses potental energy and gains kinetic energy. When it hits the floor it's velocity (ideally) remains the same in magnitude but reverses in direction. It starts moving up again.....\n\nWork is the process of conversion of energy from one for to the other. So, work is being done as the ball bounces up and down.
595852
It would end up as heat. At least that is the usual place you say all the energy went when you can't otherwise account for it. Like where does the energy go when you drop a bean bag on the ground? Yell out loud? Turn on a light? slow your fall with a parachute? And on and on and on.
526447
I actually had my class conduct an experiment on this very thing.  The problem was: does the percentage of energy lost from a ball bouncing increase, decrease, or stay the same as the height increases.  This is akin to your question, because the height at which it bounces back to depends on how much energy is left after the bounce.  \n\nWe found that the percantage of energy lost stays a constant, which means the the answer to your question is yes.\n\nFor example:\nIf you start with a height of 10 meters and drop a 1.0 kg ball, it will have 98 Joules of potential energy (PE = mgh).  If it loses 10% of its energy on the bounce, it will have 88.2 Joules of kinetic energy, which will translate to a height of 9.0 meters.  \n\nIf you double the original height, you will have 196 Joules of energy.  After losing 10% from the bounce, you are left with 176.4 Joules, which translates to a height of 18 meters after the bounce.
955845
The object has kinetic energy (1/2)*m*v^2 when first at x=0. The work friction does consumes all that energy as the spring compresses and pushes the object back to x=0. Let the distance the spring is compressed be x1. Then \n(1/2)*m*v^2 = Ff*2*x1 = mu*m*g*2*x1\n\nWhen the spring is compressed to x1, the friction has decreased the energy in the system by half and the remaining half is all stored in the spring compression because the object has come to a stop - about to start the rebound. So\nFspring*x1 = k*x1^2 = (1/4)m*v^2\nIf you solve the 1st equation for x1 and substitute the result into the 2nd, you should get the required result.
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!
915353
To answer your question simply and directly; yes. An object's motional energy loss is proportional to itself; it's a result of the forces of gravity acting directly on objects based on their respective masses and distances.\n\nAs the stone goes up in the air, it loses kinetic energy due to the pull of gravity -- it decelerates until it stops moving upward, then begins moving downward. \n\nThe second you let go of the rock, its acceleration is constant; its kinetic energy decreases as it slows down and then increases again as it falls to earth.\n\nThe kinetic energy of the rock turns into potential energy as it gets higher; kinetic energy being equal to one-half the rock's mass times its velocity squared, and potential energy being the rock's mass times the rock's height above ground times the constant acceleration due to gravity.\n\nIn a vacuum, all the rock's kinetic and potential energy would add up to the same value at any point in its rise and fall; if you threw it hard enough to give it, say, a hundred kilojoules of kinetic energy, when it reached the top of its trajectory and began to fall down, it would have zero kinetic energy but a hundred kilojoules of potential energy. On earth, it wouldn't work like that; some small amount of energy would be lost due to wind resistance, etc.
168727
Sort of.\n\nIf you drop an apple then the kinetic energy of the apple has when it hits the ground is proportional the Earth's gravitational acceleration (9.8 m/s^2). \n\nSo if you're on a different planet the "g" for that planet will be different and the kinetic energy of the apple is different, by a proportional amount.\n\nThe same is true of gravitational potential energy.
128500
The object will follow a parabolic trajectory to the earth, as observed from outside the plane.  On the plane, it will appear to fall straight down in your (moving) frame of reference.  Air resistance will retard this motion, but it's magnitude depends on the drag coefficient of the object dropped and the medium in which it is falling.