The machine which is 60%.
efficient 10kg body through
distance and spends 200 J of energy
The body is
then released on
reaching the
the ground, the KE of
body will be:
Answers
Answered by
0
given efficiency of the machine is 75% and it is spending 12 J of energy to lift that mass
according to the law of conservation of energy the stored potential energy in the mass will be converted to its kinetic energy. hence,
(12×75)/100=(m×v
2
)/2
V=
18
m/s
Answered by
2
Solution :-
As per the given data ,
- Efficiency = 60 %
- PE = 200 J
➽ Actual PE = 200 x 60 / 100
➽ Actual PE = 200 x 0.6
➽ Actual PE = 120 J
As no external force is acting on the body we can say that the total energy of the system remains conserved
By applying law of conservation of energy ,
➽ Actual PE = Gain in KE
Hence ,
➽ KE = 120 J
The kinetic energy of the body is 120 J
Similar questions