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a lift of mass 1000kg moves upward with a max mass of 800kg. it experiences constant frictional force of 4000N . if the lift moves with a constant speed of 3m/s then find the power supplied by the motor to move the lift?

7 years ago


Answers : (2)


The lift is not accelarating.

Therefore, the net force on the lift must be zero.

Force acting on the lift are:

1. Gravitational force mg on lift due to its own mass. = 1000*9.8 = 9800 N (downwards)

2. Gravitational force mg due to 800 kg load = 800*9.8 = 7840 N (downwards)

3. Frictional foce 4000 N (downwards, since lift is moving upwards and frictional force always opposes relative motion)

4. Force exerted by the rope (Tension). This force will be supplied by the motor, and this force has to be upwards. Lets call it T.

So, since net force = 0

9800+7840+4000 - T = 0

Hence, T = Force supplied by motor = 9800+7840+4000

So, T = 21640 N

Now, Power = Work/Time = Force * distance/time

or, since distance/time = velocity,

Power = Force * velocity

Therefore, Power supplied by motor = T * v = 21640 * 3 = 64920 W = 64.92 kW

7 years ago


Frictional force = 4000 N

Gravitational force = (1000 + 800 kg) x 9.8    (Force = mass x acceleration and here accelaration is due to gravity g)

                           = 17640 N

 As the lift is moving with constant speed, this means that the force provided by the lift is just as much is needed to overcome the frictional and gravitational force.


Hence, Force provided by the lift = 17640 + 4000 N = 21640 N


Now, Power = Force x Speed

                  = 21640 x 3 W

                  = 64920 W

7 years ago

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