Calculating Energy: 5.8Kg Body from 2.3m/s to 1.5m/s

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To calculate the energy derived from slowing a 5.8 kg body from 2.3 m/s to 1.5 m/s, the change in kinetic energy formula is essential. The kinetic energy (KE) is calculated using the equation KE = (1/2) * m * v^2. The initial kinetic energy at 2.3 m/s and the final kinetic energy at 1.5 m/s must be determined to find the energy change. The difference between these two values will yield the energy derived from the deceleration. Proper application of this formula will provide the correct energy calculation.
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Homework Statement


How much energy could be derived from slowing a body of mass 5.8Kg from 2.3m/s to 1.5m/s?


Homework Equations





The Attempt at a Solution


I don't even know what formula to use here so if someone could give me a couple of formlula that would be good to start with. thanks
 
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You could find the change in kinetic energy (energy of motion). Ke = (1/2)*m*v^2
 
I have tried that and it doesn't work for me??
 
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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