How Can a Block Stay on a Moving Cart Without Falling?

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AI Thread Summary
The discussion revolves around two physics problems involving motion and forces. In the first problem, a 2kg block is placed against a moving cart, and the challenge is to determine the frictional force when the cart accelerates at 32.27 m/s², which is double the minimum acceleration needed to prevent the block from falling. The second problem involves an Atwood's machine with two masses, where the goal is to find the masses given their speed, kinetic energy, and distance moved. Participants express difficulty in deriving the correct equations to solve for the unknowns, particularly in relating acceleration to the second problem. The conversation highlights the need for additional equations to solve the system of equations effectively.
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Homework Statement



1.Lou bets an innocent stranger that he can place a 2kg block against the side of a cart, and that the block will not fall to the ground, even though Low will use no hooks, ropes, fasteners, magnets, glue, or adhensives of any kind. When the stranger accepts the bet, Lou begins to push the car to the right. The cooefficient of static friction between the block and the cart is 0.6.

Find the force of the friction on the block if a=32.27 m/s2 (It is double the minimum acceleration for the block will not fall to the ground).

2. A simple Atwood's machine uses two masses, m1 and m2. Starting from rest, the speed of the two masses is 4 m/s at the end of 3.0s. At that instant, the kinetic energy of the system is 80 J and each mass has moved a distance of 6m. Determine values of m1 and m2

Homework Equations



Fs= coeff * Fn[/SUB

U=mgy.

The Attempt at a Solution



1.So I found the normal force and time by the coeff of friction. how wever I got 130 N which is the wrong answer. What did I do wrong here ??

2. So I did the U for m1 and m2 and get.

U1=m1g*6
U2=m2g*-6

Then I add 2 equations together and get m1-m2= 80/6g

I know that I need one more equation since I have 2 unknowns but I cannot think of what I can do with the speed and time info. Can anyone help me ?
 
Last edited:
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Does acceleration play any part in number 2 ??
 
Has anyone found out anything yet ?

I was still stuck at figuring out the 2nd equation relating to the given speed and time.
 
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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