Calculating Work and Power in Mechanical Systems

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To calculate the power required to raise water 3 meters and discharge it at 10 m/s, the power is determined to be 238.2 W using the formula for power as the rate of doing work. For the second scenario involving a 1 kg mass and a 3 kg mass connected by a pulley, the total work done by the frictional force, given a resistance of 0.5 N, is calculated to be 3.8625 J. Key equations include kinetic energy (KE = 1/2mv^2), potential energy (PE = mgh), and the relationship between work and energy. Understanding these formulas is essential for solving problems in mechanics. Mastery of these concepts is crucial for first-year physics students.
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1. Find the power required to raise water 3m vertically from a tank and discharge it through a nozzle of cross-sectional area 0.0003m^2 at a speed of 10m/s.
2. A mass of 1kg rests on a horizontal table. It is connected by a taut, light, inextensible string passing over a smooth fixed pulley, at the edge of the table, to another mass of 3kg hanging vertically. The 3kg mass it at a height of 0.5 metres above an inelastic floor when the system is released from rest. There is also a frictional resistance to the motion from the table of 0.5 N. Determine the total work done by the frictional force, assuming that the 1kg mass stop before reaching the pulley.
Your help would be greatly appreciated.
 
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I forgot to say, these are the answers:

1. 238.2 W
2. 3.8625 J

I've just no idea how to get them :-(
 
well I am in my first year of physics... but what is the equation of power... and then break down work into its equation... this is just a hing to get it started... if I am worng I am sorry just trying to help ;-)
 
KE = 1/2mv^2
PE = mgh
Work done = change in energy = force x displacement
Power = rate of doing work = force x velocity
F = ma
It has something to do with the above formulae, but no matter how i use them I can't get the right answer.
 
Last edited:
Please help 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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