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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A father racing his son has half the kinetic energy of the son, who has half the mass of the father. The father speeds up by 1.0m/sand then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son?

A fully loaded, slow-moving freight elevator has a cab with a total mass of1200 kg, which is required to travel upward 54 m in 3.0 min , starting and ending at rest. The elevator’s counterweight has a mass of only 950 kg, and so the elevator motor must help. What average power is required of the force the motor exerts on the cab via the cable?

(a) at a certain instant, a particle-like object is acted on by a force F=(4.0N)i^-(2.0N)j^+(9.0N)k^ while the object’s velocity is v=-(2.0m/s)i^+(4.0m/s)k^. What is the instantaneous rate at which the force does work on the object?

(b) At some other time, the velocity consists of only a y component. If the force is unchanged and the instantaneous power is -12W , what is the velocity of the object?

A cord is used to vertically lower an initially stationary block of mass M at a constant downward acceleration ofg4 .When the block has fallen a distance d , find

(a) the work done by the cord’s force on the block,

(b) the work done by the gravitational force on the block,

(c) the kinetic energy of the block, and

(d) the speed of the block.

A 0.30 kgladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (k=500N/m) whose other end is fixed. The ladle has a kinetic energy of 10 J as it passes through its equilibrium position (the point at which the spring force is zero). (a) At what rate is the spring doing work on the ladle as the ladle passes through its equilibrium position? (b) At what rate is the spring doing work on the ladle when the spring is compressed 0.10 mand the ladle is moving away from the equilibrium position?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free