Understanding Force and Mass in Kangaroo Jumps

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The discussion revolves around solving a physics problem involving a kangaroo's jump and the effect of carrying a baby kangaroo on its height. A 40-kg kangaroo can jump 2.0 m without a baby but only 1.8 m with one, prompting the question of the baby kangaroo's mass. Participants discuss the difficulty in applying free body diagrams and the relevant forces at play. The original poster expresses confusion over the calculations and the physics concepts involved, particularly in determining the relationship between force, mass, and height. The conversation highlights the importance of understanding the forces acting on the kangaroo during its jump to solve the problem effectively.
danield
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Hey.. i am having problems solving the following question...

2) A 40-kg kangaroo exerts a constant force on the ground in the firs 60 cm of her jump, and rises 2.0 m higher. When she carries a baby kangaroo in her pouch, she cna rise only 1.8 m higher, What is the mass of the baby kangaroo?

I am not interested in the answer so much.. but in the logic of the problems.. i just can't do a free body diagram for them..
please help
,dan
 
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Well, for (1), which forces are acting on the crate? Further on, what does the constant velocity tell you about the forces acting on the body?
 
ok 1.. i could solve.. i used the equation of force applied+weight=force of friction, and i found force applied and the direction.
But I am still having trouble wiht 2)
 
Well I've been trying to do something like 2y/g=t^2, then V=y/t and then A=v/t, but i guess that is not working, i think I've been seeing it from the wrong angle, but any way i try i can't get the answer :(
 
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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