The two masses in the figure are released from rest. after the $3$ kg mass has fallen $1.5$ m, it's moving with speed of $3.8$ m/sec. what is the change in mechanical energy done on the system during this time interval by the frictional force on the $2$ kg mass.

enter image description here

My trial is the the change in mechanical energy is the change in Kinetic energy since the potential is constant as the 2Kg mass moves horizontally.

So $$W=\Delta K.E. + \Delta K.E._{\text{ext}}$$ where $W$ is the work. $K.E.$ is the Kinetic energy due to tension force $K.E._{\text{ext}}$ is loss due to friction.

I don't know if this works but it gives me a wrong answer at the end (compared to the solution given). I wonder how to solve it?

  • 2
    $\begingroup$ You just defined homework-like problem... $\endgroup$
    – CoilKid
    Commented Dec 31, 2015 at 20:26
  • $\begingroup$ The phrase "change in mechanical energy done on the system ... by the frictional force" is extremely difficult to parse. Energy isn't "done" on a system - work is done. There is a loss of potential energy of the system (3 kg drops), and increase in kinetic energy (motion of 2 and 3 kg mass); the difference is lost due to friction. But how that maps to your question is something that was lost in translation (for me, at least). $\endgroup$
    – Floris
    Commented Jan 1, 2016 at 2:51
  • 1
    $\begingroup$ Please note that Physics.StackExchange is not a homework help site. Please read this Meta post on asking homework-like questions and this Meta post for "check my work" problems. $\endgroup$
    – Kyle Kanos
    Commented Jan 1, 2016 at 12:22

1 Answer 1


The change in mechanical energy done by friction will be same as work done by friction. Now there are two ways of solving this problem. You can simply calculate the work done by friction using $$W= Fx$$

force of friction will be constant (since the friction is kinetic) and you are provided with the distance moved so work done by friction can be easily calculated.The second method is you can conserve total mechanical energy of the system. $$\mathbf{U_{net~final}} - \mathbf{U_{net~initial}} = \mathbf{W_{friction}}$$ U is mechanical energy. This energy includes the net kinetic energy of system as well as its net potential energy. This is basically same as your method but you did not take potential energy into account. Also this statement of your question is incorrect

KEext is due to tension force.

Please notice that the kinetic energy of object is due to its motion with respect to a reference frame. And because this motion is decided from all of the forces acting on it we should use categorize KE by objects and their motion not by tension and friction force.

  • $\begingroup$ I neglected potential energy since it's the same before and after the motion since it's on the same height. So the difference would be zero. Isn't this true? $\endgroup$
    – FNH
    Commented Jan 1, 2016 at 11:05
  • 1
    $\begingroup$ Yeah that's true but the block hanging there will move and thus its potential energy wil change. $\endgroup$
    – Jatin
    Commented Jan 1, 2016 at 11:18
  • $\begingroup$ Oh, I thought that I would treat each body seperately as a system. thanx for clarifying $\endgroup$
    – FNH
    Commented Jan 1, 2016 at 18:08

Not the answer you're looking for? Browse other questions tagged or ask your own question.