WHy is work done by the force of gravitation?
102640
work is done to change the height of an object.  If the object is always the same distance above the center of mass, then there is no change in height in which case no work is done.  In an elliptical orbit the distance between the objects changes so work is done.
942036
When I push the wall, the wall pushes me back. Otherwise i'd have gone thru the wall.\n\nBy the way, 'work' in physics does'nt deal with its literary meaning.
997785
All mass produces gravity.  Ever massful particle, every atom exerts some gravitational force on other massful particles.\n\nPhysicists are still trying to produce the details how gravitational fields behave, theories of gravitons (the particles when mediate the field), etc.\n\nA short description of theories and current thinking is available here at Wikipedia:\nhttp://en.wikipedia.org/wiki/Gravity\n\nAnother good source of information would be to read "A Briefer History of Time" by Stephen Hawking.
231563
gravitational potential energy increases with altitued and the stregth of gravity\n\n\n\nwhile work is done, potential energy is converted into kinetic
56969
the work-energy theorem says that work is equivalent to energy.\n\nso, if you lift a 1 kg object 1 meter, you do some amount of work against gravity.  if you do the math, you'll find that the work is W = 9.8 kg*m^2/s^s.\n\nthe work-energy theorem says that this is equivalent to a change in energy, so delta_E = W = 9.8 Joules.\n\nbasically it is a simple statement of the conservation of energy, where a change in the potential energy equals the work put into the system.\n\nnote that it only applies for conservative forces, like gravity or the electrostatic repulsion.  if you are pushing something against friction, then there is an additional loss of energy to heat at the frictional surface, and the work-energy theorem does not apply.
51910
Weight is a force, but when something is weighed you actually measure it's mass assuming you are in an environment in which the force of gravity is constant (i.e., you're at a fixed location <radius from the center> on a planet or massive planetary mass).\n\nof course, one's mass can be approximated by W=mg at fixed radii from such planetary masses (like the earth, a massive asteroid, a superdense star, whatever...) if the force of gravity, g, is known at that radius.\n\nThere are 2 possible variables here - mass and the force of gravity.  Which one do you know?  usually an approximation for g is given.  Which one do you measure, then...that would be mass.
127000
This problem is solved with the application of work and energy principle which states that total work done on a moving particle is equal to the change of its kinematics energy:\nW=T2-T1\nHere the only work done on the particle is dot to gravitation force and is equal to -m*g*h:\n\n-m*10*0.45 = (1/2)*m*(1)^2-(1/2)*m*(v)^2 --->\nm*v^2=(4.5+0.5)*m*2=10*m --->\nv=sqrt(10) m/s\n\nv is equal to sqare root of 10.(approximately 3.162 m/s)
755944
Net force is required to change the motion of an object, or accelerate the object. Often forces are balanced, so net force = 0, so motion does not change.\n\nWork is in the same units as energy, not force. Work is related to force by \nW = F*d*cos(theta) where F is the applied force, d is the distance through which the force has been applied, and theta is the angle (if any) between the force and the distance. \n\nFor example if a 100 Newton vertical force is applied to lift an object to a table 5 meters above the starting point, 500 Newton*meters, or 500 joules, of work have been done.
509011
WORK:\nwork is said to be done when force is applied on a body and the body covers some distance in the direction of the applied force.\n\n                          W=F*S,WHERE:\nF=force applied\nS=distance\nNOW EXAMINING UR QUES\nif u lift an object in the direction of the force applied ,then yes work is done and if u lift it 3 times along the direction of applied force then work is tripled.\n\nONE SITUATION CAN BE THERE:\nif work done when force is not along the direction of motion-\nlet the angle of work be cos(theta)\n        cos(theta)0 =1\n        cos(  "  )30=(root)3/9\n        cos(  "  )40=1/(root)2\n        cos(  "  )60=1/2\n        cos(  "  )90=0
216326
Work done along a vertical plane has to factor in gravity as a downward acceleration force of 9.81m/s2, however, is unaffected by the contact frictional force of the plane surface. Horizontal work is unaffected by gravity (as the two forces are perpendicular) however is affected by the contact frictional forces of the plane the body is moving upon which is a function of its weight & the frictional constant of the two surfaces.
1173137
Positive work is done when the displacement is in the direction of the force applied. \n\nNegative work is done when the displacment is in the direction opposite to the force applied.\n\nZero work is done when there is no dispalcement, even i some effort is applied-eg. just holding a mass. \n\nst
57362
Weight is a unit measure of gravitational force.
503357
Maybe it's not really a mass generated force at all. It could just be that it is masses effect on space-time. Actually, I guess it could be that it's not the earth for example that causes gravity but space-times effort to put matter in clumps and roll it into a ball. The bigger the mass the more space-time there is to push against it. Einstein's example was to imagine a bowling ball sitting on a large rubber sheet. The bowling ball would deform the sheet and the curvature of the rubber represented gravity in space-time. The closer you get to the ball the more curvature you have. That would explain why gravity is weaker with added distance.
734494
(a) Work done by gravity = change in potential energy of the book. Now change in potential energy = m*g*h2 - m*g*h1=2*10*(10-1.5) =  170J. So, work done by gravity is 170J\n(b)Change in gravitational potential energy = work done = 170J as explained above.\n(c) Potential energy when textbook is released = m*g*h1 = 2*10*10=200J\n(d) Potential energy when it reaches the hands = m*g*h2 = 2*10*1.5=30J\n(e) If U =100J at ground level, work done by gravity = 170J, the work done does not change with change in reference.\n(f) change in potential energy also remains 170J\n(g) U at release point = 100 + 2*10*10 = 300J\n(h) U at hands = 100 + 2*10*1.5 = 130J
650418
Work is when you exert a force and an object moves in the direction of the force.  Power is the rate at which work is done.  And please don't copy this answer for homework or something like that because that would be cheating.
941702
Sometimes you ask thought-provoking questions and sometimes you ask rather elementary questions like this one. I think you already know the answer, unless it's asked on an unstated level (e.g., having to do with relativity or cosmology). Anyway, the work, or energy, is the time integral of the power. If power is constant, you can say it's the product of power and time of application.
13005
Its because physics seems to be really confused about this issue. We're all taught that Gravity is a force of attraction between masses. However, when you look at General Relativity you find that it is not. It is the warping of space-time by mass that causes masses to follow curved geometries that make it look like there is a force acting between them. Where it is really confusing is that then they say that Gravity is transmitted through gravitons - which seems counter intuitive to the whole GR scene. I think that's why Physics is still struggling to find a GUT (Grand Unified Theory) - they are trying to shoe-horn gravity into a place it doesn't belong.