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.
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.
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.
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.
bravo...!!
ReplyDeletePlease tell me if a coolie carrying 60kg on his head & climbs a staircase, what is the work done
DeleteIf the coolie is just climbing the stair and we assume that he leaves the stuff there itself then there will be two work done. One due to gravitation which can be found using the equation W = mgh. And the frictional work done which is w = mu.mg.s where s is the horizontal movement and h is the vertical movement.
Delete