Ad or subtract friction when a force is applied ?

The question goes. Calculate the work done on an object with a mass of 20kg to be moved horizontally if a force of 10N is applied at an angle of 30degrees of the horisontal plane. if 1 there is no frictional force and 2 if there is a frictional force with magnitude 5N.
1)
W=F x S so
W=10cos30 x 10
W=86.6 joule
2)
Here is where I am confused
If understand the formulas correctly is should be
W=F x S
W=(10cos30 - 5)x10
W=36.6 joule

This does not seem right. I want to add the 5N friction force the time taken will obviously be longer if there is a friction but the work should also be more. Unless I am trying to calculate it to simply.

1 Answer
Jun 14, 2018

I do not think that with friction the Work diminishes...otherwise you could reach a point where with a lot of friction you need no Work to move the object!

Explanation:

The idea is to look at your problem from a visual/intuitive point of view:
consider the Egyptians pushing a big stone block to build a pyramid. They put cylindrical logs in between the block and the ground to reduce friction and so reduce the fatigue!!! This fatigue is the Work (in simplest terms).

Basically, in your system (you+block) the energy goes from you to the block and this moving energy is called Work. Now, without friction Work it is easier while with friction is harder because you need to use more energy to heat the floor or to make that screechy sound you have when you push the block across the floor!
The energy transferred (Work) will be greater.
So you will get:

#"Work"_"(total)"="Work of the Force"+"Contribution due to Friction"#

[Friction is not conservative so we cannot say Friction Work!!!]

Giving:

#"Work"_"(total)"=136.6J#