# Confusion relating the application of Work-Energy theorem

1. Could someone kindly explain the solution to my confusion? When we walk at a constant speed up a steep hill, is work being done?

2. Then why isn't our kinetic energy changing as work done is equal to change in kinetic energy according to work-energy theorem? (Isn't this contradicting work energy theorem?)

• At constant speed, the net force is... – Danu Aug 20 '14 at 18:12
• You are confusing the net work with the work done by the person walking up the hill. There is also work done by gravity, and these two works together are the net work. – jwimberley Aug 20 '14 at 18:16
• Comment to the question (v3): Echoing @jwimberley, note that the definition of work depends on which force is considered. E.g. the work in the work-energy theorem refers to the work of the net force. – Qmechanic Aug 20 '14 at 18:28

The work done by a force $\mathbf{F}_1$ when a particle travels a path $\gamma : [a,b]\subset \mathbb{R} \to \mathbb{R}^3$ is defined by

$$W(\mathbf{F}_1, \gamma) = \int_\gamma \mathbf{F}_1 = \int_a^b \mathbf{F}_1(\gamma(t)) \cdot \gamma'(t) dt$$

I wrote that way just as a way to make clear that the work depends both on the path and on the force. If another force $\mathbf{F}_2$ acts on the same particle traveling the same path you can calculate $W(\mathbf{F}_2, \gamma)$ and you get another work.

Forces are additive, so that if you have forces $\mathbf{F}_1,\dots,\mathbf{F}_k$ acting on a particle, you can add them up to get a net force $\mathbf{F} = \sum_i \mathbf{F}_i$. You can see that because we defined work the way we did, it is also addtive, in other words:

$$W(\mathbf{F},\gamma)=\int_\gamma \sum_i \mathbf{F}_i=\sum_i \int_\gamma \mathbf{F}_i = \sum_i W(\mathbf{F}_i,\gamma)$$

Now, since the person walks on constant speed we know that this means that the net force $\mathbf{F}$ is zero. This force is exactly $\mathbf{F}_1 + \mathbf{F}_2$ where $\mathbf{F}_1$ is the force of gravity and $\mathbf{F}_2$ the force done by the person to overcome gravity. In that case $\mathbf{F}_1 = -\mathbf{F}_2$ and you have

$$W(\mathbf{F}_1, \gamma) = - W(\mathbf{F}_2, \gamma)$$

The work done by the net force, however, is the sum of those and is zero because the net force is zero, but this doesn't imply that the work done by individual forces is zero.

The point is that the Work-Energy theorem relates the work done by the net force to change in Kinectic energy and not the work done by individual works. Looking at the proof you see the use of $\mathbf{F} = m\mathbb{a}$ and this by Newton's second law is only valid when $\mathbf{F}$ is the net force.

EDIT: as asked in comment, in simpler terms you have to understand that work only is the change in kinectic energy when the force considered is the net force. In your case, there is really work being done by the force of gravity and by the force the person does to overcome gravity independently, but the work done by the net force however, is zero since the person travels at constant speed and the net force is zero.

• It's a good idea to match the level of technicality of the question being asked, if possible. – Señor O Aug 20 '14 at 18:32
• @SeñorO, I added an edit trying to explain the same thing without refering to line integrals and so on. – user1620696 Aug 20 '14 at 18:37
• @SeñorO Not necessarily. This site is for all users who may view the question, not just the asker. +1 for this answer. – Danu Aug 20 '14 at 22:24

Since $W = F*d$, you have to be doing work, as the force of friction is being applied over a distance. The reason why your speed is not changing is because the work is mostly being used to increase your gravitational potential energy instead of kinetic energy.

1) Yes, work is being done by your muscles.

2) Work is equal to change in Kinetic, Potential, and Heat energy:

Kinetic - No change.

Potential - Increase in gravitational potential energy accounts for some of the work

Heat - Friction in muscles/bones, shoes on the ground, and in the air resistance all
produce heat, which is where all the work that didn't go to potential energy goes.