Physics Help Forum Potential energy

 Energy and Work Energy and Work Physics Help Forum

 May 4th 2016, 11:13 AM #1 Junior Member   Join Date: Jul 2013 Posts: 11 Potential energy My textbook states that if someone raises a mass vertically upward at constant velocity, the chemical energy of the body is transferred to the gravitational potential energy of the mass. Since the upward force exerted by the arm balancing the weight of the mass is doing positive work on the mass, and the weight of the mass is doing negative work on the mass at the same time, the net work done on the mass should be zero. However, the fact is that energy is stored in the mass as gravitational potential energy. Is there any contradiction in the statement?
 May 4th 2016, 12:36 PM #2 Physics Team     Join Date: Jun 2010 Location: Naperville, IL USA Posts: 2,271 No contradiction, just a misstatement of who is doing work here. Work is done by the person on the mass, not by the mass on itself. Work done by an outside entity on a system (in this case the mass) results in an increase in energy of that system. If work is done by the mass on itself then the total energy of the mass must remain unchanged. For example when a bird flaps its wings to rise off the ground it converts some of the chemical energy in its muscles into potential energy against gravity. But here the mass is inert, it has no internal energy source to expend, so it does no work on itself.
 May 4th 2016, 01:11 PM #3 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 708 Let us call the work done by the gravitational force Wg and the work done by all other forces Wo. If the upwards displacement is from H1 to H2 whre H2>H1 the gravitational work is Wg = -mg(H2-H1) The minus sign because it is opposite to the direction of the gravitational force. The total work is equal to the change in kinetic energy dE (I know there is none in this case but bear with me) dE = Wo + Wg 0.5mV1^2 - 0.5mV2^2 = Wo - mg(H2-H1) rearranging Wo = 0.5(mV1^2 - mV2^2) + mg(H2-H1) In our case since there is no acceleration V1 = V2 so 0.5(mV1^2 - mV2^2) = 0 Wo = mg(H2-H1) If we count our height from H1, that is make (H2-H1) = the height the object is lifted to we get the statement from your book = H Wo = mgH Wo is, as Chip says, the work input by your friend in using up some of her chemical energy. mgH is defined as the potential energy of the object with reference to H1 as the base. Potential Energy always has to have a reference base Kinetic Energy does not.
 May 4th 2016, 08:43 PM #4 Junior Member   Join Date: Jul 2013 Posts: 11 Could I say the gravitational potential energy is stored in the earth rather than the mass itself, as it is the earth exerts an gravitational force on the mass and doing negative work done on the mass during the lift. Energy is taken away from the mass to the earth.
 May 5th 2016, 12:50 AM #5 Junior Member   Join Date: Jul 2013 Posts: 11 I have done some research on the internet and found that the net work done on an object is related to the change in kinetic energy of the object only, but not other kinds of energy. So whenever an object is raised against gravity, the applied force on the object has already done positive work on it to increase its gravitational potential energy, similar to the case of elastic potential energy if an elastic object is stretched or bent, or compressed. Therefore, the effect of work done on an object can be individual, except kinetic energy of an object which depends on the net work done on it. Is my explanation correct?

 Tags energy, potential