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Calculation of Power and Work Done by a Lift With Example

Previously a 5 years student of science I am quite fascinated about its advancements and like to learn more and more.

example-of-calculation-of-power-and-work-done-by-a-lift

Definition of Work in Physics

The physics is a science of calculations. Everything needs to theoretically quantified before being applied to some practical situation. The physics also deals with the calculation of work done by the machine and the power it consumes in doing it. Before we start the calculation of these terms we will have to take a look at the definition.

The definition of work in physics is as follows:

"The work may be defined as the change in the energy of a physical object that occurs upon applying some force on it"

or

"The work may also be defined as the energy consumed by an object while moving upto certain distance upon the application of force on it"


Mathematically, the work is the product of the force applied and the distance it covers under the effect of this force.


So we can say

Work = Force x Distance


Unit of measurement of work

In the SI system, the work is measured in Joule. Here is the definition:


The work done will be one joule when a force of 1 Newton displaces an object by 1 metre


Mathematically,

1 Joule = 1 Newton x 1 Metre


Definition of Power

In physics the power is defined as follows:

"The power is the rate of doing the work"

So Mathematically we can say that


Power = Work done ÷ Time taken to do that work

In simple terms, the power of a machine doing some work depends upon how large work it can complete in possibly the minimum time.

So we can say


Pmax = Wmax ÷ Tmin


Where P stands for power, W work and T for time.

Unit of Measurement of Power

In the SI system, the unit of measurement of power is Watt. Here is the definition:

"The power of a machine will be one watt if it completes 1 joule work in 1 second"

Mathematically,

1 Watt = 1 Joule ÷ 1 Second

Work done by the Lift

A lifting system elevates the physical objects of some considerable mass by some height. It has to do significant work against the force of gravity to complete the task. Here we will study the theory behind the working of lift.


Let us consider the working of a lift which is lifting the mass m by the height h in the vertically upward direction. The various forces acting on mass m will be as follows:


  • Force of gravity W = m x g acting in the vertically downward direction as shown in figure

where g is the acceleration due to gravity(9.8 m/s2)

  • The pull in the upward direction by the lift equipment = P
  • The force of friction in the direction opposite to the motion of object = S


example-of-calculation-of-power-and-work-done-by-a-lift

From the image, we can understand that the forces S and W are acting in the same direction and P in the opposite. Therefore P will have to overcome both the S and W in magnitude for lifting the mass m.

Mathematically, the equation for the force required will be:


P = S+W+F

Where F is the force required to move the object.


S = μP ( μ is the coefficient of friction)

W = mg

F = m x a ( a is the acceleration with which the mass m is moving upward)


So the total force required will be:


P = μP+mg+ma

The force F will act for small distance d1 until the object attains velocity v and then it will move with the constant speed. So the acceleration will be 0, hence the force F will also become zero

example-of-calculation-of-power-and-work-done-by-a-lift

From the above-mentioned formula,

Work = Force x distance

So the work done by the lift:

H = P x Height of lifting

= (μP+mg+ma)d1 + (μP+mg)d2

Because the force F(ma) will be acting up to the height d1

Calculation of Power:

The power is the rate of doing the work, so

Power = H/t where t is the time take to do the work.

Power = {(μP+mg+ma)d1 + (μP+mg)d2}/t


When the object is moving downwards

When the object is moving downwards, the force of gravity will pull the object in the downward direction. Now the force of friction and upward pull will be required to balance it so that the object could move with a constant speed

So the equation will be:

P+S = W

P = W - S

P = mg - μP

example-of-calculation-of-power-and-work-done-by-a-lift

The equation for work done by the lift in bringing down the object will be:

Work = P x d

= (mg - μP)d

If the lift moves the object by distance d in time t seconds, the power of the lift will be:

Power = (mg - μP)d ÷ t

Problem Solving

Problem: An elevator installed in a building lifts 5 persons on average. The weight of each person is 60 kg on average. The height of the building is 10 metre and it completes one-way transition in 10 seconds. Calculate the work done and power consumed by the lift in one cycle(Given the coefficient of friction μ = 0.10)


Solution:


The mass of the object m = 60 x 5 = 300 kg

Total hieght of lifting d = 10 metre

From the formula of work,

The work done by the lift in upward motion will be = (μP+mg+ma)d1 + (μP+mg)d2

For 1 metre the object will be accelerated to 1 metre/second2 and attain the velocity 1 metre/second

so d1 = 1 metre

and d2 = 10 - 1 = 9 metre

a = 1 metre/second2

Initially,

P = mg+ma = 300 x 9.8 + 300 x 1 = 2940 + 300 = 3240 Newton

Substituting the values in above equation we get,

Work = (0.10 x 3240 + 3240)1 + (0.10 x 3240 + 300 x 9.8)9

= 3564 + 29376 = 32940 Joule


The lift will move at a velocity of 1 metre/second so the time taken to complete the height of the building(10 metres) will be = 10 seconds

We know that,

Power = Work ÷ time

Substituting the values we get

Power = 32940 Joule ÷ 10 seconds

= 3294 Watt


While bringing the mass downwards the equation for the work will be:

Work = (mg - μP)d

Here P = mg

Substituting the value of P

Work = (300 x 9.8 - 0.10 x 300 x 9.8)10

= 2646 x 10 = 26460 Joule


Power = Work ÷ Time

= 26460 ÷ 10 = 2646 Watt


So we can say that the work done by the lift in one cycle = 32940 + 26460

Work = 59400 Joule

Power of the lift required to complete one cycle = 3294 + 2646

Power = 5940 Watt


This content is accurate and true to the best of the author’s knowledge and is not meant to substitute for formal and individualized advice from a qualified professional.

© 2019 Sourav Rana

Comments

Sourav Rana (author) on April 15, 2020:

Thanks for your valuable feedback. I will be keen to know about your lessons also.

Umesh Chandra Bhatt from Kharghar, Navi Mumbai, India on April 15, 2020:

You have explained it very nicely. I have also interest in Physics and am writing a series of lessons here. Be in touch.

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