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The loaded cab of an elevator has a mass of 3.0103kgand moves 210 mup the shaft in 23 sat constant speed. At what average rate does the force from the cable do work on the cab?

Short Answer

Expert verified

The rate of doing work on the cab by the force is2.7105Wor270kW.

Step by step solution

01

Given

The mass of elevator is, m=3.0103kg.

The distance is, d=210m.

The time is, t=23s.

02

Concept

The rate at which force does the work on the object is called as power due to the force. The work done on a particle by a constant force during its displacement is given as

W=F.d

Formula:

W=F.dPavg=Wt

03

Calculate the work done

The cab of the elevator moves with constant speed. So the cab is not accelerated; hence, the forces acting on the cab are balanced. The forces acting on the cab arethetension in the cable holding the cab and the gravitational force. Hence work done by the cable is

W=F.d=TensiondisplacementW=mgd

Substitute all the value in the above equation.

W=3.0103kg9.8m/s2210m=6.2106J

The power due to force is the rate at which force does the work on the object.

So,

Pavg=Wt

Substitute all the value in the above equation.

Pavg=6.2106J23s=2.7105W

The rate of doing work on the cab by the force is calculated to be 270 kW.

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