Question

A lit of mass 200 kg moves upward with uniform velocity of 4 m/s. If the efficiency of motor is 70 % the input power of the motor is

Answer 1

Answer:

11.2 kW

Explanation:Take g = 9.81 m/s²

Output power = F v = m g v = 200 × 9.81 × 4 W

Input power = 200 × 9.81 × 4 ÷ 70% = 11200 W = 11.2 kW

Answer 2

The input power of the motor is

Mass of the lit, m = 200 kg

Velocity of this, v = 4 m/s

Output power of the motor = $\frac{Work done}{Time}$  

                                            = $\frac{F.s}{t}$  [ Work done, W = F.s ]

                                           = F.v  [∵ s/t =v ]

                                           = m.g.v

                                          = 200 × 9.8 × 4 J

                                          = 7840 J

                                          = 7.840 KJ

We know, efficiency of motor = output power/input power

                                 or, Input power = output power/efficiency

                            or, Input power =  7.840/0.7 KJ [As given, efficiency = 70%]

                           or, Input power = 11.2 KJ

The input power is 11.2 KJ.