1000 kW of power is supplied to a motor. 90% of this is transmitted to a machine operating at 80% efficiency. The machine lifts an object of 10-tons with a velocity _____ m/s. (Assume g = 10m/s2)
Answer & explanation
Correct answer: option 2
Power transmitted to machine = \(\frac{90}{100}\) x (1000) kW
= 900 kW
Power used to lift object = \(\frac{80}{100}\)x 900 KW
= 720 kW
720000 = F x v
= (\(\frac{1}{2}\) x m x v2) x v
= \(\frac{10000}{2}\) x v3
v3 = 216
\(\Rightarrow\) v = 6 m/s.