How Does Air Force Affect Kinetic Energy in a Fancart?

AI Thread Summary
The discussion focuses on calculating the change in momentum and kinetic energy of a fancart subjected to a constant force from a fan. The change in momentum was determined to be <-0.6, 0, 0> kg·m/s. For kinetic energy, the user attempted calculations using the formulas K = 1/2mv² and K = p²/2m but arrived at different results of 0.196 J and 0.225 J, respectively. To find the change in kinetic energy, the user is advised to calculate the kinetic energy before and after the force is applied and then take the difference. The conversation emphasizes the importance of correctly applying the formulas to achieve accurate results.
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Homework Statement


A fancart of mass 0.8 kg initially has a velocity of < 0.7, 0, 0 > m/s. Then the fan is turned on, and the air exerts a constant force of < -0.4, 0, 0 > N on the cart for 1.5 seconds.

What is the change in momentum of the fancart over this 1.5 second interval?

What is the change in kinetic energy of the fancart over this 1.5 second interval?


Homework Equations


K = 1/2mv2
K = p2/2m

The Attempt at a Solution


I got the change in momentum. It's <-.6,0,0>
I can't find the change in kinetic energy.
I tried K = 1/2mv2 and got 0.196
I also tried p2/2m and got 0.225
 
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You must find the KE before and after. Take the difference to get the change.
 
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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