# A man carries an object in two different ways. In which will work done be greater? [closed]

The man is carrying a mass of 15 kg over a distance of 2 meters. I think that in case(i) the work done would be greater as both the force applied and displacement are in same direction. But the answer says case(ii). How will that be possible. Please clarify. • Check for gravitational force in case (ii) – Dr_Paradox Mar 26 at 7:41
• How would you calculate the work done in (i)? – Bob D Mar 26 at 7:50
• There is no prior effort. What is and did you apply the definition of work? What did you find and what is your refined question? Remember, this site is not for doing your homework. – my2cts Mar 26 at 9:13

Rather than thinking of the force applied by the man and the distance moved by the block, think of the energy difference.

You will notice that in the first case, the body is just shifted parallel to the horizontal surface i.e. its P.E does not increase. But in the second case, the potential energy increases, which invariably means that he has done work.

Now coming to your doubt, in the first case, the man exerts a force in the upward direction (to hold the weight) while displacement is perpendicular.

In the second case, the man exerts a force on the rope, and rope transfers the force to the block, in its direction of motion (upwards).

Hope you understand

Note that the question is about the work done on the weight, not the work done by the man. The work done by the man will be greater than the work done on the weight because the man's muscles are not 100% efficient and because energy is lost due to friction.

In case (i), if the weight is carried at a constant height about the ground, then the force required to support the weight does no work on the weight because it is perpendicular to the direction of motion. And the force in the direction of motion can be made as small as you like if the weight is moved very slowly. Therefore the work done on the weight can be made as small as you like - imagine the weight is on a frictionless table and is given a small push so that it slides slowly from one point ton the other.

On the other hand in case (ii) work is done on the weight to raise it from the ground. If we assume gravity is $$10$$ metres per second squared then the work required to raise a $$15$$ kg weight by $$2m$$ is $$mgh = 15 \times 2 \times 10 = 300$$ Joules.

Therefore the work done on the weight is greater in case (ii) than in case (i).

• "Note that the question is about the work done on the weight, not the work done by the man." Sorry, but I don't see where that's made explicit in the OP's question. OTOH, the problem doesn't have a clear solution if we don't make that assumption. – PM 2Ring Mar 26 at 10:56
• @PM2Ring I agree, that is not made explicit in the problem statement, which is probably one of the things that confused the questioner. As you say, I think we must assume this was the intention of the problem to make any progress. – gandalf61 Mar 26 at 12:21

I believe there’s insufficient information to calculate the work done in (i). You need to know the velocity of the mass at 2 meters. Then, assuming it started at rest and its height didn’t change, you could calculate the net work done on the mass by the man using the work energy theorem which states the net work done on an object equals its change in kinetic energy.

Hope this helps

• The question isn't asking to calculate work done exactly it's asking in which case it will be higher – Priyanshu Kalal Mar 26 at 8:28
• Well, how do you know which is greater unless you know how much work is done in (i)? – Bob D Mar 26 at 8:33
• Yes, your question is right. But I have assumed that the total force that the man is applying in both the cases is same. If I don't assume that, it can also be possible that force in first case be very high and displacement be same so work done in (I) is high. – Priyanshu Kalal Mar 26 at 8:39
• We can assume that the man is moving without any increase in his speed. So, After assuming thus, we can say that work done against conservative forces in the first case will be tending to zero – Krishna Mar 26 at 9:19
• If we are to make any assumptions they should be consistent. For example, if we say the man pulling the weight starts at rest and ends at rest at 2 meters, we can say without doubt he has done work equal to $mgh$ since the change in kinetic energy is zero. To be consistent we could then say the man carrying the weight starts at rest and ends at rest at 2 meters. The change in kinetic energy is zero and therefore the net work done is zero. Since there is no change in height, the change in potential energy is also zero. – Bob D Mar 26 at 14:29

(I) In first case the total force applied by man has two components ,that is , a horizontal force and a vertical force. Vertical force is doing no work as there is no displacement in vertical direction. But the vertical force is to be applied because of gravity. The horizontal force is doing work. So, W= F(horizontal) . S

(ii) In this case the total force is doing the work of displacing the object. So here, the force applied by the man to overcome gravity is also doing work. Thats why , W = F(total). S

Note that in both cases displacement is same as the man is walking the same length. But the component of force in the direction of displacement is different .