State and prove the law of conservation of energy in case
of freely falling body. Find the total energy of a body of 5 kg
mass, which is at a height of 10 m from the earth and falling
downwards with a velocity of 20 m s _2 .( g=10 m s _2 )
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Answer:
Suppose a body of mass m is falling under accelerating due to gravity g from height h.
Energy of a body before free fall (Potential energy i.e. P.E.)=mgh .
Initially velocity u=0. Therefore kinetic energy (K.E)=0
Total energy =P.E.+K.E.=mgh
Now suppose it falls upto height x from top.
Potential energy =mg(h−x)
since the body is now at height h−x from ground
velocity acquired v
2
=u
2
+2gx=2gx
Therefore K.E.=
2
mv
2
=mgx
P.E.+K.E.=mg(h−x)+mgx=mgh.
Therefore energy is conserved.
Here m=600kg,h=25m,v=50m/s,g=10m/s
2
, t=60s.
Total work done =mgh+
2
mv
2
=600(10×25+
2
50
2
)=600×1500
Power required =
time
Total work done
=
60
600×1500
=15000watts=15kW
PLEASE MARK BRAINIEST
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