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Q.State and prove law of conservation of energy in case of a freely falling body. A pump is required to lift 600 kg of water per minute from a well of 25 m deep and to eject it with a speed of 50 ms^-1. Calculate the power required to perform the above task.

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2020Subjective· 8mImportance★★★★★
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For a freely falling body, KE gained equals PE lost at every point, so total mechanical energy stays constant. For the pump, power = mass flow rate × (energy given to the water per unit mass, from both lifting it and giving it speed) = 14.95 kW.

Part 1 — Conservation of energy for a freely falling body:

Consider a body of mass mm dropped from height hh above the ground, at a point where it has fallen a distance xx (so it is at height (h−x)(h-x) above the ground) and has speed vv.

Using v2=2gxv^2 = 2gx (from kinematics, starting from rest):

KE=12mv2=12m(2gx)=mgxKE = \frac{1}{2}mv^2 = \frac{1}{2}m(2gx) = mgx

Potential energy at that point (taking the ground as reference):

PE=mg(h−x)PE = mg(h-x)

Total mechanical energy at that point:

E=KE+PE=mgx+mg(h−x)=mghE = KE + PE = mgx + mg(h-x) = mgh

This is exactly the total energy the body had at the start (all PE, since v=0v=0 there): Einitial=mgh+0=mghE_{initial} = mgh + 0 = mgh.

Since E=mghE = mgh at every point during the fall (independent of xx), the total mechanical energy remains constant throughout the fall — this proves conservation of energy for a freely falling body: PE lost = KE gained, at every instant.

Part 2 — Power required by the pump:

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