Objects of different masses, same speeds.

[MDJH]

Let's say two objects of unequal masses were accelerated to the same velocity on separate flat, horizontal surfaces with the same coefficient of kinetic friction. Assume neither object was lifted off the ground and that neither went through a surface or over its edge. Which object would be the first to come to a stop?

 

My tentative guess for this is that they'd come to a stop at the same time. Here's why I guess that.

 

The forces on each object include forces of gravity, friction, and the normal force; force of gravity would be directly proportional to mass, normal force would be identical to force of gravity (given the lack of vertical motion) and friction would be the product of normal force and the coefficient of kinetic friction.

 

Since friction is the only unbalanced force, it would be the only one responsible for acceleration of the object. Friction would be proportional to normal force, which is equal to gravitational force, which is proportional to mass. Therefore, frictional force would be, by extension, proportional to mass.

 

However, acceleration would be given by the ratio of force divided by mass. Since frictional force is proportional to mass, the ratio of frictional force divided by mass would cancel out any differences in mass regardless of the actual mass of the object. Therefore, without knowing the difference in mass, one can safely conclude they take equally long to stop.

 

Did I miss anything? o.o

[swansont]

You should be able to show this mathematically, i.e. put these thoughts into equation form.

[MDJH]

You should be able to show this mathematically, i.e. put these thoughts into equation form.

Fair enough. Let's say M represents mass, A represents acceleration, G represents gravitational accelration, U represents frictional coefficient, (same for both objects) and Fg, Fn and Ff represent gravitational, normal, and frictional forces respectively. Each term will be followed by a number that refers to the object.

 

For example, assume object 1 has k times as much mass as object 2. Therefore, M1 = k*M2

 

Per object, Fg and Fn cancel out, hence A = (Ff / M) for each object.

 

However, Ff = U*Fn, which is mathematically equivalent to Ff = U*Fg. Seeing as how Fg = M*G, we can assume that (Ff1 = U*M1*G) and (Ff2 = U*M2*G)

 

Therefore, A1 = (U*M1*G)/(M1) and A2 = (U*M2*G)/(M2)

 

Therefore, because mass cancels out in this case, the differences in mass cancel, and they therefore accelerate at the same rate, thus take equally long to stop. Did I miss anything?

[swansont]

Looks good.

 

A slightly shorter path would have been to solve it only once; since the mass term cancels you get a general expression applicable for all objects, and if m did not cancel, you would know that the accelerations were unequal. But those are style points, not affecting the correctness of the answer.

[ewmon]

Friction Factor — The lighter object would stop sooner due to the effect of normal pressure on the friction factor. Friction factor tends to increase with a decrease in normal pressure (Ref: Simulation of Materials Processing). Assume a cubic ‘cow’ (). Normal pressure is proportional to mass/surface area. Mass is proportional to volume (s³), and surface area is s², and the normal pressure (mass/area) ~ s³/s² ~ s. With a decreased size, normal pressure also decreases, so the friction factor increases, and the smaller object experiences more deceleration due to friction and stops sooner.

 

Aerodynamic drag — If this occurs under ambient conditions, including atmosphere, the lighter object would also stop sooner due to the greater aerodynamic drag proportional to its mass. Again, assume a cubic ‘cow’. Drag is proportional to cross-sectional area (s²), and mass is proportional to volume (s³). The deceleration due to aerodynamic drag is a = Fd/m ~ s²/s³ = 1/s. Thus, the smaller object experiences a greater proportional aerodynamic drag and deceleration and stops sooner.

[J.C.MacSwell]

Friction Factor — The lighter object would stop sooner due to the effect of normal pressure on the friction factor. Friction factor tends to increase with a decrease in normal pressure (Ref: Simulation of Materials Processing). Assume a cubic ‘cow’ (). Normal pressure is proportional to mass/surface area. Mass is proportional to volume (s³), and surface area is s², and the normal pressure (mass/area) ~ s³/s² ~ s. With a decreased size, normal pressure also decreases, so the friction factor increases, and the smaller object experiences more deceleration due to friction and stops sooner.

 

Aerodynamic drag — If this occurs under ambient conditions, including atmosphere, the lighter object would also stop sooner due to the greater aerodynamic drag proportional to its mass. Again, assume a cubic ‘cow’. Drag is proportional to cross-sectional area (s²), and mass is proportional to volume (s³). The deceleration due to aerodynamic drag is a = Fd/m ~ s²/s³ = 1/s. Thus, the smaller object experiences a greater proportional aerodynamic drag and deceleration and stops sooner.

 

You may want to re-think your assumptions.

 

The OP stated that the friction coefficient would be the same in each case.

[ewmon]

You may want to re-think your assumptions.

 

The OP stated that the friction coefficient would be the same in each case.

Thanks, JC. The OP led me to wonder how one determines that the coefficients of friction are equal. If determined theoretically (for example, both objects are steel sliding on a steel surface), then, as I noted, the theory does not consider the influence of the normal pressure, and the coefficients are not equal. If determined empirically, then the observations themselves (for example, both objects stopped at the same time/distance) may already answer the OP's question, which is then moot.