Friction Question: Finding Time and Distance with Kinetic Coefficients

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A block of mass 2.20 kg is placed on a larger block of mass 6.90 kg, with a coefficient of kinetic friction of 0.310. A horizontal force of 19.20 N is applied to the smaller block, and the distance it travels is 3.60 m. The correct time for the smaller block to reach the right side of the larger block is 1.24 seconds, and the distance the larger block moves is 0.740 m. Participants in the discussion sought clarification on the calculations and shared their initial incorrect answers before arriving at the correct ones.
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Homework Statement



A block of mass m = 2.20 kg rests on the left edge of a block of larger mass M = 6.90 kg. The coefficient of kinetic friction between the two blocks is 0.310, and the surface on which the 6.90 kg block rests is frictionless. A constant horizontal force of magnitude F = 19.20 N is applied to the 2.20 kg block, setting it in motion as shown in the figure, part (a) . If the distance L that the leading edge of the smaller block travels on the larger block is 3.60 m, how long will it take before this block reaches the right side of the 6.90 kg block, as shown in the figure, part (b)? Use g=9.81 m/s2 as usual.

<img src ="http://capa.mcgill.ca/res/mcgill/dcmcgill/oldproblems/mcgilllib/Kilfoil/graphics/p5_65.jpg">

How far does the 6.90 kg block move in the process?




Homework Equations



Ffriction = (mu)Fn
F = ma



The Attempt at a Solution




I got 1.13s, but it says it's wrong

second one..well i need first solution to solve..

I got .548 m with 1.13s...
 
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Oops, I got the answer..kk FYI answers were 1.24 s and .740 m.
 
Can you show step by step of how you did the question?: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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