A lift of mass 200kg moves upward with a uniform velocity of 4 m/s . if efficiency of motor is 70 then input power of motor is?

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

since the lift of mass 200kg moves upward with a uniform velocity of 4 m/s , the acceleration is 0 m/s²

Newton's II law: F̂ᵣₑ = mâ = F̂ᶰᵉᵗ = 0 = F - 200g

=> upward force provided = F = 200(9.81) = 1962 N

in 1 s, the lift moves 4 m. => W = work done/s = 1962 x 4 = 7848 watts = power

efficiency = 70%, => (0.70)Pᵢ = 7848 watts

=> Pᵢ = input Power = 7848 / 0.70 = 11211.43 watts ~= 11.211 kW


hope this helps