A $60HP$ electric motor lifts an elevator with a maximum total load capacity of $2000kg$. If the frictional force on the elevator is $4000N$, the speed of the elevator at full load is close to (Given $1HP=746W$, $g=10m/s^2$)

Q. A $60HP$ electric motor lifts an elevator having a maximum total load capacity of $2000kg$. If the frictional force on the elevator is $4000N$, the speed of the elevator at full load is close to $(1HP=7464W,G=10m/s^2)$

Q. A $60HP$ electric motor lifts an elevator with a maximum total load capacity of $2000kg$. If the frictional force on the elevator is $4000N$, the speed of the elevator at full load is close to (Given $1HP=746W$, $g=10m/s^2$)

Q. An elevator can carry a maximum load of $1800kg(elevator+passengers)$ is moving up with a constant speed of $2m{s}^{-1}$. The frictional force opposing the motion is $4000N$. What is minimum power delivered by the motor to the elevator?

Q. An elevator in a building can carry a maximum of $10$ persons, with the average mass of each person being $68kg$. The mass of the elevator itself is $920kg$ and it moves with a constant speed $3m/s$. The frictional force opposing the motion is $6000N$. If the elevator is moving up with its full capacity, the power delivered by the motor to the elevator $(g=10m/s^2)$ must be at least

Q. An elevator can carry a maximum load of 1800 kg (elevator + passengers ) is moving up with a constant speed of 2 m/s. The frictional force opposing the motion is 400 N. Determine the minimum power delivered by