The Coolie Work Done Problem

The Problem

In schoool level science boooks it is often stated that coolies (porters) who carry your baggages on their heads scienntifically don't do any work. The question is, Is it reallly so? The answer is both YES and NO.
Let us see what the real answer is to this intriguing problem.

The Solution

Let us start by understanding the concept of work done.
Let us say that force acting on an object in such a way that it is moving in the same direction as the force itself.

Now the mathematical equation for work done is given as
W = F . S = F.S. cos(theta) where (theta) is the angle between force and displacement.

So in the above case force and displacement are in the same direction which means (theta) = 0 degree and hence cos(theta) = 1. Hence work done W = FS which will be meximum for the given situation.

In the above figure the force applied is at 90 degree to the displacement. In this situation (theta) = 90 and cos(theta) = 0. Hence work done W = FSx0 = 0. Hence no work done.

Now the situation for a coolie under the action of gravity terms is exactly like the second case. The cooolies has to apply force to push the weight (Force) in the upward direction and has to move (displacement) in the perpendicular direction. Hence his work done is zero.
So under the condition that the collie is moving horizontally only his work done would come out to be zero.

But wait !!!!

Don't rush to tell a coolie that he is not doing any work for he actually is doing.
The total work done by a coolie is as follows.

Work (coolie) = Work (gravitation) + Work (friction)

That means there are two aspects of the work done by the coolie. The work (gravitation) as we saw can be zero. So everything boils down to

Work(coolie) = Work(friction)

As we know friction is a non-conservative force. Which means it depends on the actual path taken. So let us try to understand the work done by a coolie through an example.

Example:

Let a coolie of mass 60 kg is carrying a load of 15 kg on his head. He goes horizontally from one place to another at a distance of 250 meters. Let the coefficient of friction between the coolie's shoes and the road  is 0.8. Now lets see what happens.

Solution

Work(coolie) = Work(gravity) + Work(friction)

Work(gravity) = F.S. cos(theta)
Since theta = 0 therefore     Work(gravity) = 0.(This is what is given in books)

Weight of coolie + Weight on his head = (60+15) kg x 10 m/sq.s = 750 N  (g = 10 m/sq.s approx)
Force of friction = mu x weight = 0.8x 750 N = 600 N approx = 600 N

This means to walk the coolie needs to exert a force of at least 600 N on the ground to move ahead.
Work(friction) = Force x Displacement
                        = 600 N x 250 m = 15000 J.

So the work done by a coolie is actually

Work(coolie) = Work(gravity + Work(friction)
                      = 0                    +  15000 J
Hence the total work done is not zero but 15000 J.

This means the books teach you only about Gravitational work done and nothing else. Hence never ever tell a coolie that he has done no work.

Life of a Huge Star v/s a Smaller Star

The Problem:

It has been found both theoretically and observationally that in comparison to smaller stars the bigger more massive ones die out earlier. Analgously it is like if two cars of same make A and B have fuels such as A has 10 liter of petrol and B has 50 liters then common sense tells us that B will go longer than A in distance but in case of stars it is exactly the opposite, which means that A will survive longer than B if A is less massive than B. The question is HOW and WHY?
So let us see logically how.

The Solution:

Let us first understand that when a star is born it is almost completely made up of  Hydrogen which sets off a fusion reaction to produce heat, light and huge outward pressure. This huge outward pressure is countered by the gravitational force inward. This gravitational force is responsible for the starting of the fusion reaction at the first place and then maintaining it.
Now let us assume two stars A and B. A starts with 100 units of Hydrogen and B with 100000 units. Automatically their masses differ by 1000 times with B being more massive. This means that the gravitational pull of A will be 1000 (approx) times less than B and hence when the respective fusion reactions begin A will burn much, much slower than B.
In a reverse way we can say that since B has a gigantic mass its gravitational force will also be massive and so will be the outward pressure due to reaction. More fuel will have to  burnt more rapidly to counter the inner gravitational pull and to keep the star stable. So that will lead to a rapid usage of fuel leading to a smaller life span in comparison to the smaller star.