Work Done on a system by an external force

[harpua09]

Homework Statement

A worker pushed a 26kg block 13m across a level floor at a constant speed, with a force directed 40 degrees below the horizontal. If the coefficient of kinetic friction between the block and floor is 0.40, what were a.) the work done by the workers force? and b.) the increase in thermal energy of the block floor system?

Homework Equations

F(friction) = coefficient of friction x normal force

E(thermal) = F(friction) x d

The Attempt at a Solution

I am stuck with being able to calculate the normal force on the block without knowing the magnitude of the applied force. I Know that that F(normal) = F(gravity) *m*a*cos(o)
I think that
F(friction) = -F(applied) ,
but i am not sure what to do with this. Does something cancel?

Thanks in advance

 

[collinsmark]

A worker pushed a 26kg block 13m across a level floor at a constant speed, with a [...]

That phrase "constant speed" turns out to be very important. What does "constant speed" imply about the block's acceleration? After realizing the implications of that, and with a couple basic laws, you'll have enough information to solve for the normal force, then the frictional force, then the rest.

I am stuck with being able to calculate the normal force on the block without knowing the magnitude of the applied force. I Know that that F(normal) = F(gravity) *m*a*cos(o)

Umm, something isn't quite right regarding your "m*a*cos(o)" term. I'm not sure where you got that. The normal force has something to do with the worker's applied force. Why not use that?

I think that
F(friction) = -F(applied) ,
but i am not sure what to do with this. Does something cancel?

Your above equation is only correct if you are limiting the "F(applied)" to the component of the overall applied force that is parallel to the force of friction, and it is only correct if nothing is accelerating. But don't despair, you're on the right track!