Calculating Work and Power: 50-N Force on a 2-kg Object

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A 50-N force acting on a 2-kg object from rest for 2 seconds results in a power output of 2500W. The discussion highlights the use of equations of motion, specifically SUVAT, to solve related problems. For the second scenario, where the object has traveled 2 meters, participants suggest using the equation V^2 = U^2 + 2as to find the velocity. This approach emphasizes the importance of selecting the appropriate equations based on whether time or displacement is the focus. Understanding these principles is crucial for accurately calculating work and power in physics problems.
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Homework Statement



65. A 50-N force is the only force on a 2-kg object that starts from rest. When the force has been
acting for 2 s the rate at which it is doing work is:
A. 75W
B. 100W
C. 1000W
D. 2500W
E. 5000W
ans: D

66. A 50-N force is the only force a 2-kg crate that starts from rest. At the instant the object has
gone 2m the rate at which the force is doing work is:
A. 2.5W
B. 25W
C. 75W
D. 100W
E. 500W

Q1 solved but i have no idea how to do the second one!
 
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Look up the standard equations of motion (SUVAT). Some are more useful for problems involving time (eg Q1) and others for displacement (Q2).

Perhaps this one...

V^2 = U^2 + 2as

where s = displacement
 
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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