If a ball is thrown into the air, we may assume it has an initial positive kinetic energy $E_k$, and its gravitational potential energy $U_g$ is 0. Once it reaches its apex, $E_k=0$ and $U_g=E_k(initial)$. Thus, the net work $W_{net}= 0$, according to the work-energy principle,

$$ W_{net}= \Delta U_g +\Delta E_k $$ However, the force of gravity performs work on the ball according to $$ W = \int_{r_0}^{r} m\mathbf{g}\cdot d\mathbf{r} $$ Where is the work to balance out the work due to gravity?


Your energy equation is missing a subscript. It should be $$W_\mathrm{net, external} = \Delta K + \Delta U$$

The work in that expression is external; its agent is outside the system, and is doing work on the system. So we need to decide what's the system so that we can declare what is internal and what is external.

If the system is the earth and the ball, then there is no external work. The work done by gravity is between two members of the system and is internal work. (In fact, the definition of potential energy is $\Delta U = -W_\mathrm{internal}$) In this case $$ 0 = \Delta K + \Delta U = \Delta K + mg\Delta y$$ On the other hand, if the system is only the ball, then gravity does external work on the system, but it comes in with a negative sign because the displacement is in the opposite direction from the force. Also, potential energy is not defined for a single particle. Potential energy is the energy associated with the configuration of two particle. It's just not defined for a single particle. So$$ -mg\Delta y = \Delta K$$ Of course we get the same answer: nature doesn't care about how we set the boundary between a system and its environment.

So in one case you have potential energy, in the other you have external work, but you don't have both at the same time.


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.