This isn't homework. I'm reviewing physics after many years of neglect. As with most of my posts, I made this problem up.

 

Let object A have mass [latex]m_A[/latex] and object B have mass [latex]m_B[/latex]. One of A's surfaces is flat, as is one of B's. These flat surfaces are in contact and slide relative to each other in a straight line for a distance, x, experiencing kinetic friction. The kinetic friction, of magnitude f, is constant and is the only force acting on A or B in the direction of motion, and the only other forces acting on A and B are the normal forces pressing the surfaces together. The normal forces are assumed to be equal and opposite, so that the only acceleration of the objects is in the direction of friction. No net force acts on the center of mass.

 

During a finite time interval, the speed of each surface relative to the other decreases from [latex]v_i[/latex] to [latex]v_f[/latex] due to friction. The nonconservative work done by each surface on the other is [latex]W_{noncons} = -fx[/latex] since each surface experiences the same displacement and the magnitude of the frictional force exerted by each surface on the other is the same since the two frictional forces comprise a force/reaction force pair.

 

Therefore the change in kinetic energy of A, [latex]\Delta K_A[/latex] as seen from B's restframe is equal to the change in kinetic energy of B, [latex]\Delta K_B[/latex], as seen from A's restframe, since we are assuming that no other forces accelerate the objects.

 

Therefore [latex]\frac{1}{2}m_A v_{f}^{2} - \frac{1}{2}m_A v_{i}^{2} = \Delta K_A = \Delta K_B = \frac{1}{2}m_B v_{f}^{2} - \frac{1}{2}m_B v_{i}^{2} \Rightarrow m_A = m_B[/latex], an absurd result.

 

Something's wrong and it's driving me crazy! Please help me find the error in this reasoning, thanks.

 

Let object A have mass and object B have mass . One of A's surfaces is flat, as is one of B's. These flat surfaces are in contact and slide relative to each other in a straight line for a distance, x, experiencing kinetic friction. The kinetic friction, of magnitude f, is constant and is the only force acting on A or B in the direction of motion, and the only other forces acting on A and B are the normal forces pressing the surfaces together. The normal forces are assumed to be equal and opposite, so that the only acceleration of the objects is in the direction of friction. No net force acts on the center of mass.

 

 

 

 

That's not possible.

 

There must be forces acting on A an B in or against the direction of motion, other than friction.

Edited January 10, 201412 yr by studiot

That's not possible.

 

There must be forces acting on A an B in or against the direction of motion, other than friction.

Why isn't it possible? If A and B have an initial relative velocity, and are slowed by friction while in contact, what other force acts in the line of motion?

Therefore the change in kinetic energy of A, [latex]\Delta K_A[/latex] as seen from B's restframe is equal to the change in kinetic energy of B, [latex]\Delta K_B[/latex], as seen from A's restframe, since we are assuming that no other forces accelerate the objects.

I'm not sure, but speaking of "A's rest frame" and B's sounds suspicious. Each changes velocity, so which rest frame are you talking about? Before, or after the acceleration? From an arbitrarily chosen rest frame, if m_A != m_B, then A and B don't have an identical change in velocity relative to the observer.

I agree with md65536. Neither A nor B remain in a single inertial frame, and KE is not invariant. That's where the trouble is.

 

Why isn't it possible? If A and B have an initial relative velocity

 

Good point.

 

Further

 

KE is not a function of relative velocity.

Edited January 10, 201412 yr by studiot

The nonconservative work done by each surface

 

When one group of molecules (solid state object A) collide with other group of molecules (solid state object B), one gives energy to the other, heating it. Conservation of energy happens at quantum scale.

 

When your hands are cold, you are starting rubbing one on the other hand, and feeling hot between them.

Edited January 11, 201412 yr by Sensei

Thanks! A and B's rest frames are not inertial.