The height at which the mass will be lifted is; 3 meters
How to utilize efficiency of a machine?
Formula for efficiency is;
η = useful output energy/input energy
We are given
η = 60% = 0.6
Input energy = 4 KJ = 4000 J
Thus;
0.6 = useful output energy/4000
useful output energy = 0.6 * 4000
useful output energy = 2400 J
Work done in lifting mass(useful output energy) = force * distance moved
Useful output energy = 800 * h
where h is height to lift mass
Thus;
800h = 2400
h = 2400/800
h = 3 meters
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