Calculating Kinetic Energy: Understanding the Equation and Common Errors

AI Thread Summary
The discussion centers on calculating the kinetic energy of a 3kg object with a velocity of < 6i -1j > m/s. The calculated kinetic energy is 55.5 J, while the book states it as 60 J, leading to confusion. Participants agree that the discrepancy likely arises from rounding practices and significant figures, noting that the velocity has only one significant figure. The consensus is that the book's answer is a rounded figure to maintain consistency with significant figures. This highlights the importance of understanding rounding in physics calculations.
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kinetic energy, help!?

A 3kg object has a velocity < 6i -1j > m/s. What is its kinetic energy at this moment? According to the book I'm using the ans is 60J but I get 55.5 J...What em I doing wrong?
 
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They probably just rounded their answer. If you use sig figs, there's only one sig fig in < 6i -1j >, so they rounded 55.5 to 60 to conserve sig figs.
 


i'm getting the same answer as you
 


Hey thanks guys, I was really going nuts here trying to figure it out, it only made sense that the answer was rounded.
 
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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