Calculating Work and Thermal Energy in a Block-Floor System

[bbbbbev]

[CLPhys1 10.P.042.] Worker Pushes Block A worker pushed a 27 kg block 6.2 m along a level floor at constant speed with a force directed 30° below the horizontal.
(a) If the coefficient of kinetic friction is 0.20, what was the work done by the worker's force?

(b) What was the increase in thermal energy of the block-floor system?

This seems like it is supposed to be really easy, but I can't get it.

For (a), I set up a force diagram, and since the block is moving at constant velocity, I knew that F_worker-->block_x = friction, so F_w-->b_x = (mu)_k(m)(g). Since W_worker_x = F_worker-->block_x*(delta x), I just multiplied everything together and then tried to find F_worker-->block by dividing F_worker-->block_x by cos(30). I tried it with cos(60), too, but neither got the right answer.

For part (b), I set W_friction = friction_k * (delta x) * cos(180), but that didn't work either.

I don't know what else to try. I would really appreciate any help anyone can give me!

Thanks
Beverly

 

[thenewbosco]

If you sum the forces in the x-direction you can get:

[tex] \sum F_{x} =0 = F cos 30 - F_{k} = 0 [/itex]

[tex] therefore, F cos 30 = \mu_{k}N [/itex]

[tex] and \F = \frac{\mu_{k} N}{cos 30} [/itex]

you should then sum the forces in the y-direction to solve for F and equate the two expressions. Now you can solve for the normal force.

then all you have to do is use the fact that:

[tex] W = F\cdot d cos \theta [/itex]

substitute in the 1st equation for F

therefore [tex] W = \frac{\mu_{k} N}{cos 30} \cdot d \cos \theta [/itex]

To find the increase in thermal energy of the floor you can use the same equation for force of kinetic friction times distance to find the work done by friction.

 

[bbbbbev]

Thanks! I had forgotten about summing the forces on the y-axis! Thanks so much!