Help me proving this Kinetic Friction Coefficient

AI Thread Summary
The discussion revolves around deriving the kinetic friction coefficient for a system involving a block on a table connected to a falling mass via a pulley. The desired expression for the kinetic friction coefficient is μ = m/M - a/g. Participants are struggling to arrive at this formula, with one user providing an alternate expression: μ = [mg - (M+m)a]/Mg. A condition is noted where if M is much greater than m, the expression simplifies to the required form. The conversation also briefly touches on unrelated topics, but the main focus remains on the physics problem at hand.
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Consider the system: A block of mass M is placed on a level table and connected to a mass m by a light string running through a pulley. If mass m is released, it will start to fall and the wooden block will be pulled across the table.

Prove that the kinetic friction coefficient can be given by the expression: μ = m/M - a/g

I have tried, but could only get this expression: μ= [mg - (M+m)a]/Mg . I can't get to the one outlined in the question. What am I missing?

Thanks in advance.
 
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Is there any condition given that M >> m . since in that case (1 + m/M) will tend to 1, and your expression will reduce to the required expression. I am also getting the same answer as you got.
 
me too :smile:
 
Does 4 make a greater charm than 3? :smile:
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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