# Work Energy Theorem Application

Does the work energy theorem

Work done by all forces = Change in Kinetic Energy


Include the work done by external torque ? and does $\delta$K include rotational kinetic energy?

Also Should we calculate net force and then calculate work done OR add the work done by each force ?

## 1 Answer

I think you should treat separately usual translational forces and torques. Indeed the work done by a torque is equal to the difference of rotational kinetic energy. To keep things simple, assuming a constant torque, you get:

$$\tau \, \varphi = \Delta K= \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2$$

where $\tau$ is the constant torque, $\varphi$ is the rotation angle, $I$ is Inertia momentum and $\omega_f$ , $\omega_i$ are final and initial angular velocities.

As far as your second point, if you carefully consider the orientations and the relative signs, the results will be the same, no matter which way you follow. For me, it's more intuitive to compute first the total net force and then compute the work.

• Could you explain it to me in more detail ? – InquisitiveMind Dec 11 '16 at 15:57
• If you push a box on an iced floor, you're doing a work on the box. The work you do on the box turns into kinetic energy of the box itself. If you want to compute the kinetic energy difference you have two ways: either you sum all the forces acting on the box and then you compute the work by doing the scalar product of the total force and the displacement vector; otherwise you compute the works done by each force acting on the box (maybe someone is positive, maybe someone is negative) and then you sum all them up. This concerns the "translational word". – AndreaPaco Dec 11 '16 at 16:02
• Thats clear, what about variable torque , do we need to integrate torque.theta and find the work done ? – InquisitiveMind Dec 11 '16 at 16:06
• If the forces you apply are such to cause a rotation, you'd better computing rotational quantites apart. My advice is to compute the total torque acting on your box and then evaluate the rotational kinetic difference with the formula I've written you in the answer. Indeed yes: you need to integrate! The simple linear formula I've shown you is valid only if the torque is constant during the rotation. – AndreaPaco Dec 11 '16 at 16:08