Solving Two Veggie Patties Physics Problem

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The problem involves two veggie patties with masses of 113 g and 139 g being pushed on a grill with a force of 5.38 x 10^-2 N. The acceleration of the system was calculated to be 0.21 N/kg. The discussion focused on determining the force exerted by each patty on the other, with confusion regarding the relevance of vertical forces due to negligible friction. Ultimately, the user resolved their misunderstanding and found the correct answers. The conversation highlighted the importance of carefully reading the problem statement to avoid misinterpretation.
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Homework Statement



Two Veggie patties, in contact with each other are being pushed across a grill. the masses are 113 g and 139 g. Friction is negligible. The applied horizontal force of magnitude is 5.38 x 10 ^-2 N is exerted on the more massive burger. Determine a) the magnitude of the acceleratio nof the two burger system and b) magnitude of the force exerted by each of the two burgers on the other.

i know someone posted this here but i checked that thread, the method is not provided there.

Homework Equations



EF = ma
Fg = mg

Answer at the back:
a = 0.21
F = 2.41 x 10^-2

The Attempt at a Solution



Found acceleration which is 0.21 N/kg.
Dunno how to find the magnitude of the force of B/a or A/B.
 
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vpv said:

Homework Equations



EF = ma
Fg = mg

You probably used the first equation to do part a. Why do you think the second equation is useful?
 
heth said:
You probably used the first equation to do part a. Why do you think the second equation is useful?

Its useful because u need to find the vertical forces.

For the upper patty, it would be:

EF = ma = 0
Fg - Fn - F bottom/top = 0
Fg = Fn + F bottom/top

I don't know if that is correct tho. Is there a normal force, as friction is 0?
 
What would be the net horizontal force on the small patty, given its acceleration and mass?
 
The net horizontal force is given. its 5.39 x 10^-2 N. There is no friction.
 
NVM i got the answer lol. i misread the question. Thanks for ur help
 
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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