A machine which is 75 percent efficient uses 12 Joules of energy in lifting up a 1 kg mass through a certain distance. The mass is then allowed to fall through that distance. What will its velocity be at the end of its fall?
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Efficiency = (Output energy / Input energy) × 100
75 = (Output energy / 12 Joule) × 100
Output energy = (75 × 12 Joule) / 100 = 9 Joule
Actual energy utilised to lift the mass up is 9 Joule
Energy required to lift the mass up = mgh
9 J = (1 kg × 10 m/s² × h)
h = 0.9 m
Final velocity of freely falling body after falling from a height “H” is
v = sqrt(2gH)
= √(2 × 10 m/s² × 0.9 m)
= √(18 m²/s²)
= 4.24 m/s
∴ Velocity at the end of its free fall is 4.24 m/s
[Note: I took g = 10 m/s² for easy calculations]
75 = (Output energy / 12 Joule) × 100
Output energy = (75 × 12 Joule) / 100 = 9 Joule
Actual energy utilised to lift the mass up is 9 Joule
Energy required to lift the mass up = mgh
9 J = (1 kg × 10 m/s² × h)
h = 0.9 m
Final velocity of freely falling body after falling from a height “H” is
v = sqrt(2gH)
= √(2 × 10 m/s² × 0.9 m)
= √(18 m²/s²)
= 4.24 m/s
∴ Velocity at the end of its free fall is 4.24 m/s
[Note: I took g = 10 m/s² for easy calculations]
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