Calculating Work and Friction in a Horizontal Surface: A Practical Example

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A person pulls a 70 N bag 200 meters at a constant speed using a force of 40 N at a 50-degree angle. The work done by the person is calculated to be 5142 J. To find the work done by friction, it is noted that since the speed is constant, the change in kinetic energy (KE) is zero. The potential energy (PE) change is also zero on a horizontal surface, leading to the conclusion that the net force is zero. This indicates that the force exerted parallel to the surface equals the force of friction.
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A person pulls her 70 N bag 200 meters at a constant speed.
The force exerted is 40 N at 50 degrees.

I can figure out the work the person does (5142 J).

But I'm having trouble figuring out the 'work done by the force of friction'

I know that Wfriction = delta KE + delta PE
...Since the speed is constant KE should be 0, but I'm not sure what to do about PE.. I know PE = mgy and delta PE = PEfinal - PEinitial but since it's a horizontal surface wouldn't that be 0? I must be doing something wrong because it can't be 0..
 
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The net force is ____ because the bag's velocity and speed are _____ which means the force by the ______ equals the force of _______.
 
Just to make sure I understand

The net force is _0___ because the bag's velocity and speed are __constant___ which means the force by the _parallel component of force(40N*cos(50))_____ equals the force of _friction______?
 
You've got it.
 
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