What constitutes work?

Question

I'm asked: "An object is moving along a floor at constant velocity. Thinking about this situation as realistically as possible, is work being done on the object?"

And the solution is: "Work is being done on the object. To move at a constant velocity, a force must be pushing the car so as to overcome friction, air resistance, etc."

However, since work is defined as $W = F · x$, and $F = ma$, wouldn't work be zero? Because acceleration is zero, and hence the sum of the forces acting on the object is zero, and hence the dot product of F and x is zero?

Or do we count work even when it is cancelled out y equal and opposite work?

Thanks!

EDIT: I can see that the frictional force is not doing any work (since the object is not moving in the direction of the friction) but that the push-force IS doing work, since the object is saying in the same direction.

However, the net force on the object is still $0$. So I understand that we can attribute work to the push-force -- but is the object experiencing any OVERALL work? How is this possible if acceleration is $0$?

Answer 1

You have to do some work on the object to keep it moving at a constant speed! Friction against the floor would stop it dissipating its kinetic energy as heat. You need to push the object applying to it a net constant force in order to keep it moving, so you do some work equals to the net force you do on the object times the path you make the body take. That is:

• Friction:$$F_{fric}=-\mu_s\cdot gm+const$$where $\mu$ is a constant $\in[0,1]$ that represent the strength of the static friction, $m$ is its mass and $g$ the acceleration of gravity. This force points against the direction of its motion.
• The work needed to take the object moving at constant speed: $$F_{tot}=0=F_{push}+F_{fric},\quad\rightarrow\quad F_{push}=\mu\cdot mg\\ W_{push}=\mu\cdot mg\cdot x=const.\cdot m\cdot x$$ The heavier the object, the more you need to do work.

Answer 2

I would like to clarify that frictional force is actually doing work, but the work is negative, i.e. energy is transferred away from the object due to the frictional force.

With that in mind, for the object to remain its velocity (thus its kinetic energy), work done is required to transfer energy into the object.

I think you're right that there is no overall work on the object (i.e. no net energy transfer into the object), but if we don't look at overall, but instead on individual force, it is indeed work done on the object.