How Is Work Calculated When Pulling a Wagon Up an Incline?

[nameVoid]

A person pulls a wagon up an incline that makes an angled of 30 deg with the horizontal with a force on the handle of 30 pounds which makes an angel of 30 deg with the incline find the work done in pulling the wagon 100 ft
20cos30 will be a force along the surface of the incline and since vectors are in the same direction force*distance 20cos30*100=1000sqrt(3) which is the same as if you were to remove the incline?

 

[Mark44]

A person pulls a wagon up an incline that makes an angled of 30 deg with the horizontal with a force on the handle of 30 pounds which makes an angel of 30 deg with the incline find the work done in pulling the wagon 100 ft
20cos30 will be a force along the surface of the incline and since vectors are in the same direction force*distance 20cos30*100=1000sqrt(3) which is the same as if you were to remove the incline?

Do you know about punctuation? Without punctuation it's much more difficult to parse what you have written.

The force applied is NOT in the same direction as the wagon is moving.

 

[nameVoid]

20lbs 30 degrees

 

[Mark44]

The pulling force is being applied at an angle (not angel) of 60 degrees to the horizontal.

 

[nameVoid]

I would also suppose taking the force vector to be <10,10sqrt(3)>=a and the distance vector <50sqrt(3),50>=b
work= a*b =500sqrt(3)+500sqrt(3)=1000sqrt(3)
but taking the force vector along the incline to be 20cos30 and the distance 100 ft along the incline gives the same results

 

[Mark44]

How did you get 20 cos(30 deg)?