# If the definition of work done is change in kinetic energy, when I am pushing a box along the floor at constant speed am I not doing any work?

And can I still say the chemical energy in my muscles is being converted to kinetic energy? Sorry if this is a stupid question, but I was doing some physics questions in a textbook, and one question was about why a Sankey diagram of energy conversions in a car didn’t show the kinetic energy, but if it’s at constant velocity why does that mean chemical energy from fuel is not being converted to kinetic energy if the car is moving? And another one asked what happens to the work done on a box when I push it with 65N 52° to the floor at constant velocity, with no distance given, and the answer states that all the work done is converted into thermal energy, why is this? Is the movement of the box not kinetic energy?

• pushing a box with a constant speed means some other force also exists which is counterbalancing your force and thus work done by your force is equal to negative work done by the other force.. Commented Jan 24, 2021 at 14:03

There is some variety in how work is defined in different textbooks.

The most common one is to define it simply as the product of force and distance (or generalizations of that, taking into account the directions of the vectors). This style of definition defines the work done by one force. There is a problem with this definition, which is that the distance is not well defined for a force like kinetic friction, where there is slipping and no well-defined point of contact. In this style of definition, there is a work-kinetic-energy theorem (not definition), which equates the total work done on an object (counting the "distance" as the distance moved by the object's c.m.) to the change in KE. In your example, your hand's force on the box is not the only force doing work.

The style of definition used by your book is actually the one I prefer. However, defining work as a change in KE doesn't quite do the job, since we want to be able to talk about the work done by individual forces. So in this style of definition, a better way to go is to define work as a transfer of energy by a macroscopic force -- any type of energy, not just KE. So in your example, as you say, the chemical energy in your muscle and liver glycogen is being transferred into the box. But at the same time, there is energy flowing out into frictional heating. The two energy flows cancel, so the box's macroscopic KE stays constant.

If you are pushing a box at constant force $$F$$ and the box is moving at constant speed (acceleration $$a=0$$) this means that the toal force $$F_T$$ acting on the box is $$F_T=0$$ (because $$F_T=ma=0$$). So there must be another force $$F_f$$ due to friction acting on the box that cancels out your force $$F_f=-F$$ (the same applies for a car, with $$F$$ the "propulsion force" and $$F_f$$ the friction with the road).

So indeed the total work $$W_T$$ exerted on the box is going to be 0 because $$W_T=\Delta K = 0$$ where $$\Delta K$$ is the change in kinetic energy and that is 0.

However one is usually interested in the amount of work one is exerting and from your point of view you are exerting a force $$F$$ by a distance $$x$$ so that your work is going to be $$W=Fx$$ You are indeed consuming energy when pushing a box or driving a car.

From the point of view of the box, on the other hand, there also is an opposite force $$F_f$$ exerting work $$W_f$$. Summing the two components

$$F_Tx = (F+F_f)x= \Delta K=0$$ indicating that $$Fx+F_fx=0$$ so that $$W=Fx=-F_f x=-W_f$$ i.e. the total work is 0 but t he individual ones ($$W$$ and $$W_f$$) are not, but they cancel each other).

So the KE theorem $$W=\Delta K$$ holds if you sum all the forces acting on the object. However, there still is kinetic energy hidden somewhere in the fact that you are exerting work: your muscles are moving, the ground is going to heat up because of friction, etc. Energy still is conserved, but your work, instead of going directly into KE of the box is going into "more degraded" kinetic energy because of friction. But you are exerting work even if the box does not accelerate!

So:

• what is the work I am exerting? $$W=Fx$$

• what is the total work exerted by external forces on the box? $$W_T=0$$

Pushing a box at a constant velocity is not increasing its KE as it is not accelerating. If there were no frictions or other forces acting on the box it would keep moving at its constant velocity without you pushing it. However the frictions with the floor and air resistance will act to resist its constant velocity so the energy you use to push it only opposes the frictional resistances and is converted to thermal energy.