Force exerted by a motor, velocity, and friction

Some motor make an object move on a surface very slowly. One can write at steady state : $$F_{motor}-F_{friction}=0$$. The friction is supposed to come from the surface so : $$F_{motor}-\mu_s m g=0$$ with $$m$$ the mass of the object.

Now I would like to know the velocity of the object. so I thought to use $$P=F_{motor}v$$, and knowing the power of the motor, $$v=\frac{P}{\mu_s m g}$$, but it looks wrong to me because it's not clear in my head wether $$F_P$$ which is in $$P=F_P v$$ should be $$F_{motor}$$ or $$F_{friction}$$. And also because that in the work I'm doing I don't get the result I'm expecting...

Could you help plz ? Is the expression $$v=\frac{P}{\mu_s m g}$$ right ?

• At constant velocity the force of the motor is equal in magnitude and opposite in direction to the force of friction. Otherwise, the object would be accelerating. – David White Jan 10 at 16:52
• I do not understand how this system works? Is there a drive shaft and wheels to move the object that slides on the ground? – Vincent Fraticelli Jan 10 at 19:26
• Does it change something ? – J.A Jan 10 at 19:38
• I think so . Because the concept of $Fmotor$ is not very clear? If there are wheels, it is the reaction of the ground on the wheels that pushes the vehicle: an action of the ground. And if the wheels do not slip, the associated power is zero! The clearest is perhaps to apply the theorem of kinetic energy: it directly involves the power of the engine (inner action) without the need for "Force engine". – Vincent Fraticelli Jan 10 at 20:02
• I think your expression is correct if the whole vehicle slides on the ground, without wheels. – Vincent Fraticelli Jan 10 at 21:13

You need to keep in mind there are two types of friction coefficients: static ($$μ_s$$) and kinetic (or sliding) friction ($$μ_k$$). The coefficient of kinetic friction is generally less than kinetic friction.

In order to get the mass moving, the motor needs to exert a force equal to the maximum static friction force, or $$F_{motor} =μ_{s}mg$$.

Once the mass starts moving the friction force opposing motion will be the kinetic friction $$μ_{k}mg$$ which will generally be less that the static friction force or,

$$μ_{k}mg<μ_{s}mg$$

Now if the motor force remains unchanged, it will be greater than the kinetic friction force and the mass will undergo a constant acceleration. So the velocity of the mass will depend on how long this situation lasts since $$v=at$$. If, at any time after the mass starts moving the applied motor force is reduced so that it equals the kinetic friction force, acceleration will cease and the mass will continue at the velocity it had just before the force reduction. So clearly if you want the mass to be moving slowly, the motor force needs to be reduced shortly after the mass starts moving.

So once the mass is moving at constant velocity, the power delivered by the motor should be

$$P=μ_{k}mgv$$

or

$$v=\frac{P}{u_{k}mg}$$

The key is the coefficient of friction when the mass is moving is the kinetic coefficient of friction.

Hope this helps.