Mechanics problem about Work

[dt4u]

OK I need some help on this problem. I know that work is only done in the X direction due to the F cos 30 component of both people, however it does say frictional surface. Doesn't the F sin 30 component (the y component) from person A increase the frictional force (Ff) in the x direction vs the person B who has no y component? Ff is going to be co-efficient of friction times Fnormal which is the weight plus the downward y component from F sin 30. At the very least I would say D, no enough information because we don't know the distance in X that the block was pushed.

Thanks

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[Nugester]

Normal force= mg + Fsine(30). F(xnet)= ma(net); with Fx(net) = Fcos(30)- Fr, with Fr= uk*normal force. Person A would be doing more work than B. This makes sense as someone pushing a box slightly into the ground with friction has a harder time moving it than someone pushing straight on. This is also the same reason why if person A was dragging the box at an angle above horizontal, they do less work as there is less friction.