General work of friction question

AI Thread Summary
The discussion centers on understanding the concept of work done by friction and energy loss in two scenarios. For object #1, the work done by friction is -4.73, while for object #2, it is -4.63, leading to confusion about which scenario represents greater work. The negative signs indicate energy consumed, with larger absolute values representing more energy loss. In terms of energy loss for the ball, scenario #1, with a final energy of 0.27, indicates a greater loss than scenario #2, which has a final energy of 0.37. Clarifying whether "greater" refers to magnitude or scalar value is essential for accurate interpretation.
jerad908
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Homework Statement
General WORK of friction question
Relevant Equations
energy
if the Work done by friction is -4.73 for object #1 but the work done by friction for object #2 is -4.63, in which scenario is the Work greater - the signs are throwing me off.

Also, let's say the initial elastic energy of a ball is 5 and its final energy is 0.27 (scenario #1) vs its final energy is 0.37 (scenario #2), in what scenario did it lose more energy? I am going to assume its scenario one because 0.27 final energy is less than 0.37 final energy however, the work done is -4.73 and isn't that smaller than -4.63?
 
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Welcome, jerad 908! :cool:

Try to see it as energy consumed (hence the negative sign) by friction between two surfaces or internally by the material of the ball (energy commonly converted to heat).
 
jerad908 said:
if the Work done by friction is -4.73 for object #1 but the work done by friction for object #2 is -4.63, in which scenario is the Work greater - the signs are throwing me off.
It depends which work you are referring to (work done by friction or work remaining in the object) and what you mean by greater than (as magnitude or as a scalar).
It might help if you post the question as given to you, word for word.
 
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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