A horizontal force of magnitude 49.8 N pushes a block of mass 4.32 kg across a floor where the coefficient of kinetic friction is 0.552. (a) How much work is done by that applied force on the block-floor system when the block slides through a displacement of 3.43 m across the floor?

Question

z4k4r1y4 · Answer

I did:
$$(F_{applied}-F_{friction})*(3.43)=91J$$

z4k4r1y4 · Answer

|dw:1448206315708:dw|

naveenbabbar · Answer

As you have mentioned that the block is moving, the work done by applied force is 
= Applied force x Displacement

naveenbabbar · Answer

What you have done is actually net work done or change in kinetic energy.

z4k4r1y4 · Answer

Ok that makes sense, thank you.

naveenbabbar · Answer

If you want to calculate work done by frictional force = Force of kinetic friction x displacement. This work done will be negative.

z4k4r1y4 · Answer

which means work done by applied force will be greater than total work done. now i understand.

naveenbabbar · Answer

Also the diagram you have drawn is only valid in case of sliding situations, but will not work in rolling situations.