1
$\begingroup$

What does positive and negative work done indicate? What I understand is that positive work done implies that energy is gained by the rigid body and negative work done implies energy is lost by the body.

For example, a box is kept on a rough table, and when I push it, the work done by the force applied increases the energy of the block and the work done against friction is energy lost by the block while moving against friction assuming there is some displacement. Is this correct? In that case, what would be the effect on the internal energy of the body?

$\endgroup$
1
$\begingroup$

Recall the Work Energy theorem,

$W_{cons.}+W_{non-cons}...=\Delta KE$

This relation tells us that whenever the net work done on a body is $+ve$, the body gains kinetic energy.

Notice that I used the term 'net' here. It doesn't matter what the work done by a particular force is, its the net which matters. A force may do less $-ve$ work and another force may do more $+ve$ work, resulting in net $+ve$ work. The system gains kinetic energy due to this work. Its a pretty neat way to develop intuition about the signs of work.

In your example, work done by you is $+ve$ and $-ve$ by friction. Since, the box is moving, the net work comes out to be $+ve$, therefore the body gains kinetic energy.

The kinetic energy of the system has nothing to do with internal energy. Internal energy consists kinetic energy of the molecules inside the system with respect to its centre of mass.

Since molecular motion is primarily a function of temperature, the internal energy is sometimes called 'thermal energy'.

Total energy of the system $(E)$ = U (Thermal energy) + $E_{Kinetic\space of\space system}$ + $E_{Potential\space of\space system}$ ....

As said before, internal energy includes the kinetic energy of the molecules with respect to the centre of mass. Slowing down the centre of mass won't have any effect on the internal energy. It remains the same.

$\endgroup$
1
$\begingroup$

This acquires all the more importance in thermodynamics. Here work done by the system is taken as positive. So that the first law becomes,

Internal energy change $=$ heat absorbed $-$ work done by the system

Work done on the system for instance in an adiabatic process $(Heat\space transfer =0)$ would therefore increase the internal energy of the system. In Mechanics on the other hand, there is a work-energy theorem. Net work done on the system increases the $KE$.

$\endgroup$
  • $\begingroup$ Oh . So if we consider a body like a piston and a gas then the work done would impact the internal energy . And If we are taking about a body , then it would impact it's kinetic energy. Is that correct ? $\endgroup$ – user51714 Sep 12 '17 at 7:22
  • $\begingroup$ Yes it is like that. $\endgroup$ – SAKhan Sep 12 '17 at 7:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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