An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of ${\rm{1}}{\rm{.5}}\,{\rm{m}}{{\rm{s}}^{ - 1}}$. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is: (${\rm{g}}\,{\rm{ = 10}}\,{\rm{m}}{{\rm{s}}^{ - 2}}$)Failed to load question data. Please try again later.
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