1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Work done and change in the internal energy

  1. Jul 3, 2013 #1
    1. The problem statement, all variables and given/known data

    A force F is applied on a block of mass M.The block is displaced through a distance "d" in the direction of the force.What is the work done by the force on the block?Does the internal energy change because of this work?

    2. Relevant equations

    3. The attempt at a solution

    The work done by the force =Fd
    By Work Kinetic energy theorem ,change in kinetic energy of the CM of the block is =(1/2)mvcm2

    The internal energy can change either by supplying heat to the system or by doing work on the system.

    Work is being done on the block .The KE of the CM of the block changes.But how does this work affects the internal energy of the block ?

    Kindly help me in clearing the doubt.
  2. jcsd
  3. Jul 3, 2013 #2
    If there is no friction between the block and the surface, then all of the energy due to the applied force will go into kinetic energy. However, if there is fraction, then work in the form of heat will be generated according to Frictional force X distance. The heat will go into the block and the surface.
  4. Jul 3, 2013 #3
    Hi barryj

    You didnt get my point.I understand that work done will change the kinetic energy.But its the Kinetic energy of the CM.How does a change in the KE of the CM reflect on the internal energy of the system(the block,in this case)
  5. Jul 3, 2013 #4
    I assume CM means center of mass? I don't think I have heard of KE of the CM or internal energy related to the CM. If you push a block, the surface where the block slides along the surface will get hot, if there is friction. This work will go into internal energy of the block as heat. The heat will not initially be distributed enenly and to find out the distribution at a certain time is a difficult boundry value problem.
  6. Jul 3, 2013 #5
    Barryj,I agree with you.

    I am not trying to relate KE of CM to internal energy.All I am saying is that the work done by the force changes the KE of the CM.On the other hand,internal energy is the measure of the average kinetic energy of the molecules.How do we analyse the change in internal energy from the perspective of the first law of thermodynamics as well as the work kinetic energy theorem?
  7. Jul 3, 2013 #6


    User Avatar
    Homework Helper

    The internal energy is the energy of random motion of the particles. The kinetic energy of the CM does not contribute to the internal energy. On the other hand, internal energy does not mean translational kinetic energy only. Rotating molecules contribute to the internal energy with both their translational and rotational energy. Vibrating molecules also contribute with elastic energy.

  8. Jul 3, 2013 #7
    Hi ehild :)

    You are absolutely right.But,where does the work done on the block go ?

    Can you give an example of a force(other than friction) which could cause a change in the internal energy of the block .
  9. Jul 3, 2013 #8
    Anything that would heat the block. Put the block in a microwave, the water in the block will heat and therefore the IE will increase
  10. Jul 3, 2013 #9


    User Avatar
    Science Advisor

    In general ##dU = \delta W + \delta Q##. In a quasi-static process ##\delta W = -pdV## where ##V## is the volume of the system and ##p## is the pressure and we can write ##dU = \delta Q - pdV##. If you apply pressure to the block then you can change its internal energy (obviously since the atoms making up the block are tightly bound, you will not really increase the volume by much when you apply pressure).
    Last edited: Jul 3, 2013
  11. Jul 3, 2013 #10

    Doc Al

    User Avatar

    Staff: Mentor

    Into the macroscopic translational KE of the block!
  12. Jul 3, 2013 #11


    User Avatar
    Homework Helper

    You know that the CM of a system of point masses moves as if all external forces acted at the CM.

    If an external force does some work on a box of gas molecules, the CM gets some velocity and the whole system gains kinetic energy equal to the work done.

    The molecules perform random motion and in case the force does not change the state of the gas, the velocity distribution of the molecules will not change. They have their random velocity vir inside the box, and the box travels with velocity V. So the i-th molecule has got the velocity vi=V+vir with respect to the ground. Assume identical particles, so the mass is m, the same for all of them. The work done is equal to the change of the KE. The total kinetic energy of the system is KE=Ʃ1/2 mvi2=1/2V2Ʃm + mVƩvir+Ʃ1/2mvir2.
    The first term on the RHS is the KE of the CM, as if the total mass was concentrated in it. The second term cancels as the molecules performs random motion, the average velocity is zero. The third term is the KE or the random motion - the same as in the box in rest. The work of the external force on the box changed the kE of the CM and left the internal energy unchanged.

  13. Jul 3, 2013 #12
    As I recall, technically, the CM is a location, not a molecule or any substance. It is the location where the mass would be located if it were in an infinitesimally small dot. How can a CM, a location, have energy?
  14. Jul 3, 2013 #13


    User Avatar
    Science Advisor

    We think of it as describing a hypothetical particle whose mass is the total mass of the aggregate system and location is at the center of mass of the system.
  15. Jul 3, 2013 #14


    User Avatar
    Science Advisor

    Strictly speaking, no one has said the CM has any energy. What you have been told is that the kinetic energy of the entire block can be calculated by treating it as if the mass were concentrated at the CM.
  16. Jul 3, 2013 #15
    In the usual form of the first law, the change in kinetic energy and potential energy of the system are assumed to be negligible. But, more generally, they should also be included on the same side of the equation as the change in internal energy.

  17. Jul 4, 2013 #16
    Thanks ehild :smile: .This is what i was looking for.
  18. Jul 4, 2013 #17


    User Avatar
    Homework Helper

    An example:
    If that block is not quite rigid, the force causes deformation when it starts to act. That deformation starts elastic waves in the material, and the elastic energy transforms to heat.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted