a shot-put of mass 4.0kg is dropped 2.5m vertically. what is the potential that is lost?
962133
E potential = mgh\n= 4 kg ( 9.81 m/s^2) ( 2.5)\n=98.1\n\nPotential energy was not lost....it transformed into Kinetic Energy\n\nKinetic + Potential = Total Energy\n\nLaw of Conservation of Energy
103276
The work done to lift the box to that height is stored in the body as the potential energy. Hence the work done is 400 J.\n\nAs for the height, potential energy is given by m*g*h, and the weight is given by m*g (=200N here).\nTherefore, 200h = 400 => h = 2m.
840035
Ok, we're going to assume the 4.6268 cm is in addition to the 58.8cm, so the ball doesn't stop until it has fallen 63.4268 cm.  Therefore it has lost .0833*.634268*9.8 = .51778 Nm of energy.\n\nSince all this energy goes into compressing the spring, and its energy is .5kx^2, k = 2*.51778/.046268^2 = 483.74 N/m
915417
Basically, in conservation of energy, you have:\n\nEnergy(initial) = Energy(final).  This becomes\n\nKE(initial) + PE(initial) = KE(final) + PE(final), neglecting air resistance.  When the projectile hits the ground, there is no potential energy because h = 0.  Therefore, the equation simplifies to:\n\nKE(i) + PE(i) = KE(f)\n(1/2)m*v_i^2 + m*g*h = (1/2)m*v_f^2.\n\nWe know that:\n\ng = 9.8 m/s^2\nv_i = 175*sin(45) m/s (because we are only taking the vertical component)\nh = 165 m\n\nThe masses cancel, now solve for v_f:\n\nv_f = sqrt([v_i*sin(45)]^2 + 2gh) = sqrt(175^2 + 2*9.8*165)\n\nSolving, you get v_f = 184 m/s\n\n------------\n\nHope this helps
916585
Conservation of momentum so mv is constant.\n\nTotal final mass = 5333+14.6 = 5347.6g\n\nTotal final momentum = 5347 x 1.7 gm/s = 9090.92 gm/s\n\nBullet's initial momentum = 9090 gm/s\n\nMomentum/mass = velocity\n\n9090.92/14.6 = 622.666m/s
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.
573522
kinetic energy is 1/2 mv² so it depends on mass and velocity
905951
First of all I assume the height of \nbasket above ground is 2+3.05=5.05\nLet v be the required velocity.\nvertical component=vsin40\n=v*.766\nfrom conservation of energy\nheight gained\n=[v*.766]^2/2g=3.05\nv^2=3.05*2*9.81/.766^2\n=101.986\nv=10.09m/sec
821414
The bullet's vertical motion is governed by the vertical forces only and not governed by horizontal and vice versa.\nIn each case the bullet's vertical motion is governed by gravity acting on itself. The gravity force gives an acceleration of 9.81 met/sec squared. Both have the same distance to go through and both have a initial vertical velocity of 0. Therefore they both reach at the same time, neglecting air friction.
815859
the vertical component of the velocity applied to the ball is given by\n\nsin 35 = velocity/24.5\n\nvertical velocity is 14.05 m/s\n\nV^2 = u^2 +2gs\n\nat the maximum height, v = 0 so\n\n0 = (14.05)^2 + 2(-9.8)s\n0 = 197.48 + -19.6s\n19.6s = 197.48\ns = 10.07 metres\n\n\nmaximum height the ball reaches is 10.07 metres...
514815
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)
772547
Increase your elevation or find a court with less gravitational force.   It's  a physics thing.
257076
dude u need to start doin your own hw. ill do it this time\n5. well you have to check all the forces on the bag. the first force is gravity, which is downwards and its 9.8m where m is the mass (Force = mass x acceleration) so gravity force is 20.5kg * 9.8 = 200.9N. then the 2nd force is the force of your own movement. you accelerate it upwards at 5m/s^2. so you apply a force of 5x20.5 = 102.5N on the bag upwards. but this is the thing. it gets confusing. the force is applied on the bag, not on the items in the bag. the items in the bag  is what  makes the bag weigh so much. so if you pull the bag up with 102.5N force, the items in the bag pull downwards with the same force (another one of newtons laws) which means the total downwards force is 102.5 + 200.9 = 303.4N downwards. which is more than 245N so the bag will rip. \n\n6. well in general F  = ma. but the F is total force. in this case, there is a friction force along with the applied force.\nso Fa + Ff = ma\n   37 + Ff = 5 x 5.8\n   Ff = 29 - 37\n   Ff = -8N     its negative because its opposing the applied force. in this case i can choose which direction i want to be positive, i chose the direction that the object is moving. you could put negatives in front of the applied force, friction force and acceleration   and youll get the same answer.  so friction force is 8N.\nb. the coefficient of friction is calculated by Ff = u x Fn\nwhere Fn is the normal force and u is the coefficient. you already know Ff. the normal force is easy its basically  the force tha topposes gravity. if the object isnt moving up or down (if its on a surface). so then gravity is 5 kg x 9.8 = 49N \nso then Ff = u x 49\n    8 = u x 49\n u = 8/49\n u = 0.1632..... coefficient of friction is 0.1632\n\n7.\na.  the force is 107N and the mass is 5.1kg. so acceleration is 20.98m/s^2\nb. so acceleration = Vf-Vi/t\n  20.98 = Vf - 0 / 10s\n  Vf = 10 x 20.98 = 209.8m/s\nfinal velocity is 209.8m/s\nc. the net force that acts on the object is only the downward gravity force which is 9.8x 5.1 = 49.98.\nd. the direction of their velocity goes down when the velocity becomes 0. so you can do this by saying acceleration is gravitational and it is -9.8. initial velocity is 209.8 and final velocity is 0. and now u gotta find time. so  a = Vf-Vi/t\n-9.8 = -209.8 / t\nt = 209.8/9.8\nt = 21.4s   so it starts going down   21.4s after 10s after its been released (the first 10 s  its accelerating upwards)   so it starts going down 31.4s after its been realeased.
657765
1. the object is pushed up by the weight of its equivalent volume of water, 4-3.2=0.8l\nSo the object's density is equal to its mass divided by its volume:\n4kg/0.8l=4kg/0.0008m3=40,000kg/8m3=5000kg/m3\n\n2. the force decreases with the square of distance, so it will be 3^2=9 times smaller, which is 6N.
466064
You will of course have more Potential Energy at the top of the stairs. PE(grav) = H x M x g, where H is your height relative to the ground, M is your mass, and g is the acceleration of gravity (approximately 10 meters persec persec on Earth). Thus the higher you are, the greater the gravitational potential energy. So as you walk down the stairs, you decreasing your height relative to the ground, and thus lowering your PE. So what happened to the potential energy? Did it just disappear? No, the potential energy got converted into kinetic energy (you walking down the stairs) and also heat.\n\nJust as an additional note: the reason you have gravitional Potential Energy at the top of the stairs is because there is a gravitional force acting on you by the Earth. Of course when you are on the ground, there is still gravitional force acting on you. So when we talk about potential energy, we are talking about the potential energy difference between the top of the stairs and the ground, not some absolute number.
438060
Without regard to air resistance you solve the problem as such.\n\nv(f) = final velocity\nv(i) = initial velocity\na = acceleration\nt = time\nx = displacement (distance)\n\nWhen you throw the ball up vertically, you know several things:\n1) v(i) = 30 m/s\n2) v(f) = 0 m/s\n3) a = -9.8 m/s^2 (this is because acceleration due to gravity goes in the opposite direction of the object)\n\nSolve for x with this equation:\nv(f)^2 = v(i)^2 + 2ax\n0^2 = (30 m/s)^2 + 2(-9.8 m/s^2)x\n0 = 900 m^2/s^2 - (19.6 m/s^2)x\n(19.6 m/s^2)x = 900 m^2/s^2\nx = (900 m^2/s^2)/(19.6 m/s^2)\nx = 45.9 m (rounded off)\n\nNow you know how high the ball went.  The time it takes to reach this is equal to the time it takes to get back down (assuming that it is the same distance up and down).\n\nTo solve for t, use this equation:\n\nv(f) = v(i) + at\n\n0 = 30 + (-9.8)t\n9.8t = 30\nt = 30/9.8\nt = 3.06 s  this is going up to max height.\n\nto check if it will be the same time going down, use this equation:\n\nx = v(i)t + (1/2)at^2\nv(i) = 0 (because when you reached max height, your velocity will be zero)\na = 9.8 because your ball is travelling in the direction of the acceleration.\n45.9 = 0 + (1/2)(9.8)t^2\nt^2 = 45.9/4.9 = 9.367\nt = 3.06 s\n\ndone
752440
The force applied to the beast (by the bullet) would have to be calculated by the angle/arc of it's descent, this new force differential may cause a new vector.  ie  If it was falling straight down and you fired straight up...no change.....if it were coming in at 45^ angle down, the bullets force will slow it slightly, and it's arc down will have increased due to the continuing force of gravity...with less momentum.  That is...also only if the head stays attached....if not then...the two bodies must be calculaated by the individual resistance/force of the bullet....    In the end...take one step to the right...after your shot...and watch out for falling bits and pieces.