Taaseen Rahman
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When calculating work done by a man, do we use the horizontal component of the force he applies or the horizontal resultant force?
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Absent Agent
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(Original post by Taaseen Rahman)
When calculating work done by a man, do we use the horizontal component of the force he applies or the horizontal resultant force?
Bumping for you.

Interesting question! I would also like to know the answer.

Edit: depending on the context, I would say both but I will explain this later on.
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Taaseen Rahman
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I think I got it, we'll use the horizontal component of force he applied, not resultant. As resultant will contain e.g backward force by friction. The force by friction is on a separate the work done by friction on the opposite direction I guess
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Konanabanana
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(Original post by Taaseen Rahman)
When calculating work done by a man, do we use the horizontal component of the force he applies or the horizontal resultant force?
The force you use is always parallel to the direction of motion. if your motion is vertical then you will use the vertical component. This should answered in the definition of work done.

The work done is the force exerted on an object parallel to the direction of that objects motion W=FD
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(Original post by Taaseen Rahman)
I think I got it, we'll use the horizontal component of force he applied, not resultant. As resultant will contain e.g backward force by friction. The force by friction is on a separate the work done by friction on the opposite direction I guess
Well, I think if you are to find the work done by the man against friction (assuming that the man has a constant velocity) then you are correct as the resultant force would be zero.

Considering the work done against gravity when an object is lifted in a gravitational field, we know that the force is the weight of the object and that the height is the distance, that is, W=F\times d=mgh. However, you should realise that work cannot be done against gravity if the applied force on the object is less than or even equal to the weight of the object, meaning that practically a greater amount of work is done but only the force equal to the weight is responsible for the work done against gravity.

As a result, if we are to find the work done by a car as it's accelerating over a particular distance, then the work done (against friction as well as the work done increasing the kinetic energy of the car) is equal to W=F_{friction+resultant} \times d.

Someone please correct me if I'm wrong.
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Taaseen Rahman
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The horizontal component was in parallel to the direction of the motion
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