# Can someone explain this equation more clearly: $\Delta E_{Total}=\Delta K + \Delta U = W_{External}$

I asked my Physics teacher and he explained it to me, but I am still confused on how to use it. $$$$\Delta E_{Total}=\Delta K + \Delta U = W_{External}$$$$

I think it's the "work-energy theorem", but whenever I google it, I end up with one that looks like this: $$$$W_{Net}= K - K_{0}$$$$

As I understand:
(Total Energy acted on an object) = (change in Kinetic Energy) + (change in Potential Energy) = (External Work)
In the problem attached, I got the answer by setting: $$$$\frac{1}{2}kx^2 = mgh$$$$ then solving for "h". I watched a video to get this solution. What I want to know is how would I get to this equation using the equation given to me by my teacher? All the homework I am doing seem to use different equations, but I think they can all be had from using this original equation.

## 3 Answers

The net work $$W_\text{net}$$ can be broken down into two pieces:

1. The work done by conservative forces (i.e. - forces that can have a potential energy associated with them), such as gravity or spring forces. Let's call this $$W_\text{conservative}$$.
2. The work done by other such forces. This is what you call $$W_\text{external}$$.

Thus, $$W_\text{net} = W_\text{conservative} + W_\text{external}.$$

You are correct that the work-energy theorem states $$W_\text{net} = K_f - K_i$$ where $$K = \frac{1}{2} mv^2$$ is the kinetic energy (this is nothing more than an algebraic rearrangement of the kinematics formula $$v_f^2 = v_i^2 + 2a(x_f - x_i)$$ paired with the defnition of net work $$W_\text{net} = \mathbf F_\text{net} \cdot \Delta \mathbf x$$). Substituting this in we have the following.

$$\Delta K = W_\text{conservative} + W_\text{external}$$

The work done by conservative forces (the energy transferred into the object by the conservative forces) is exactly the amount of energy lost from the potential energy (this is basically the definition of potential energy. As an equation, this is $$W_\text{conservative} = - \Delta U$$ where $$U$$ represents kinetic energy. Substituting this fact in and then rearranging gives the following.

$$\Delta K = - \Delta U + W_\text{external}$$

$$\Delta K + \Delta U = W_\text{external}$$

The quantity $$K + U$$ is what we as physicists define as total mechanical energy, or $$E_\text{total} := K+U$$. This allows us to write the following.

$$\Delta E_\text{total} = \Delta K + \Delta U = W_\text{external}$$

So, in short, yes, this is just another form of the work-energy theorem. It's just written in such a way so that $$W_\text{external}$$ means all the work done by forces acting on the object that do not have a potential energy $$U$$ associated with them.

• "this is nothing more than an algebraic rearrangement of the kinematics formula...". If you are going to give the background, might as well also state that Newton's second law is invoked here too. Good answer! – Aaron Stevens Oct 3 '18 at 3:40

$$W_{External}$$ is the work done by some extra force (different from gravity and the spring). There is no such force here, so $$W_{External} = 0$$. I think this should already point you in the right direction to solve the problem. You're also moving from a stationary state to another stationary state, so there's also no difference in kinetic energy. Therefore $$\Delta U = 0$$. Your potential energy is the sum of the spring's elastic energy ($$\frac{1}{2}kx^2$$) and the gravitational potential energy ($$mgh$$). However, their "deltas" have different signs, because one is decreasing and the other is increasing (energy is transferred from the elastic potential to the gravitational potential): therefore we should consider them with different signs. From that, you get to the equation you correctly identified and solved.

The equation you found on Google doesn't use the concept of potential energy. Recall that the work exerted by a force is $$W_f = -\Delta U_f$$. The equation you found puts everything in $$W_{Net}$$, not only "external" forces, but also the ones that could be expressed as potential energy.

The equation is a conservation of energy equation for a system (though missing heat, $$Q$$, if this was a thermodynamics problem). Consider the cheese and spring together to be the system. A system can contain two types of energy, kinetic and/or potential energy. The total energy of the system $$E_{total}$$ is the sum of its kinetic and potential energies.

In this case there is no net transfer of energy (work or heat) between the system and its surroundings. If there were there would be an increase or decrease of the total kinetic and/or potential energy of the system. In this case all of the initial potential energy of the spring is transferred to gravitational potential energy of the cheese. When the cheese falls back and compresses the spring to its initial deformation, the potential energy of the cheese is transferred back to the spring. In other words, all energy is exchanged within the system between the spring and the cheese. Therefore if we consider $$W_{ext}$$ to be the net energy transfer between the system and its surroundings in the form of work, $$W_{ext}=0$$.

Thus

$$\Delta E_{total} =\Delta K + \Delta U = 0$$

Finally, neither the cheese nor spring is in motion relative to a stationary frame of reference (floor supporting spring, for example) initially or finally when the cheese stops at height h, so there is no change in kinetic energy of the system and $$\Delta K=0$$.

So what we are left with is

$$\Delta U = \Delta U_{spring} + \Delta U_{cheese} = 0$$

$$-\frac {kx^2}{2} + mgh=0$$

Solve for $$h$$

Hope this helps