# How do I found the Kinetic Energy of a mass being lifted?

I did an experiment where I lifted a mass and measured the force. I know the change in height, the mass, and the work of the pull. How can I use the work-kinetic energy theorem to find the final and initial kinetic energies? I don't use $\frac{1}{2}mv^2$, do I? I would think initial and final velocities would be 0, so I'm having a hard time understanding the difference between the initial kinetic energy and the final kinetic energy. Can I use work of the pull and the change in height to find the kinetic energies?

• May you be talking about the potential energy instead? Then it will have some sense. – m0nhawk Oct 5 '15 at 13:43
• I don't think so. "What are the initial and final kinetic energies of the block?" Then it states the Work-Kinetic Energy Theorem. – Brendan Oct 5 '15 at 13:45
• kinetic energy gained / lost is the difference between work done and potential energy gained. – Floris Oct 5 '15 at 14:24
• In my opinion, the Work-Kinetic Energy theorem is at best useless, and at worst dangerously misleading. Beginners usually ignore the condition that there are no other forces other than the one doing the work. In your case, there is the force of your pull and gravity. The theorem does not apply! – garyp Oct 5 '15 at 14:25

The kinetic energy of the mass being lifted can be found with the equation $\frac{1}{2}mv^2$. This applies for the instantaneous velocity while the mass is being lifted. When the velocity is $0$, when you begin lifting and when you stop lifted, the kinetic energy is $0$. This is because $\frac{1}{2}m(0)^2 = 0$. The energy that you expended from your muscles transforms into kinetic energy as the mass is being lifted, and as the mass stops being lifted the kinetic energy is transformed into potential energy because the mass is in a gravitational field. Note that while the mass has kinetic energy, as it is being lifted, the kinetic energy is already being transformed into potential energy, not just when you stop lifting. The work-kinetic energy theorem does not apply to this situation because gravity is also acting on the mass. Finally, as a side note, the potential energy of the mass can be found with the expression $mgh$, where $m$ is the mass, $g$ is gravitational acceleration ($9.81 \,\mathrm{m/s^2}$), and $h$ is the height. I hope this answers your question.