Physics, asked by Rubi01072001, 5 months ago


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 Anonymous
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 Atαrαh
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