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Chapter 7: Q75P (page 176)

What is the power of the force required to move a4500kgelevator cab with a load of 1800 kgupward at constant speed 3.80 m/s?

Short Answer

Expert verified

The power of the force to move an 4500 kg elevator cab with a load of 1800 kg upward at a constant speed 3.80 m/s is 235 KW

Step by step solution

01

Given information

It is given that,

i) Mass of elevator cab 4500 kg

ii) Mass of load 1800 KG

iii) Speed v=3.80 m/s

02

Determining the concept

Using the formula of power in terms of force and velocity, find the power of force when it moves an object with a constant velocity.

Formula is as follow:

p=Fv

Where,

F is force, vis velocity and Pis the power.

03

Determining the power of the force

Now,

p=Fv

But,the total force is,

Ftotal=massofelevatorcab+massofload9.8m/s2Ftotal=4500kg+1800kg9.8m/s2Ftotal=61740N

Therefore,

P=61740N3.80msP=234612W=235kW

Therefore, thepower of the force to move an 4500kg elevator cab with a load of 1800 kg upward at a constant speed 3.80 m/s is 235kW

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Most popular questions from this chapter

To pull a 50 kgcrate across a horizontal frictionless floor, a worker applies a force of 210 N, directed20掳above the horizontal. As the crate moves 3.0 m, what work is done on the crate by (a) the worker鈥檚 force, (b) the gravitational force, and (c) the normal force? (d) What is the total work?

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