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)

6

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.