Midterm Review: Solving for Friction Work on a Rough Incline

[Sny]

Doing some midterm review and I got stuck on this question. Starting from rest, a 5.0 kg block slides 2.1 m down a rough 30.0° incline. The coefficent of kinetic friction between the block and the incline is µk = 0.436. Determine the work done by the friction force between block and incline.

My first thought was mgµk times distance, since work equals force times distance, but that didn't work. I'm not sure if the fact that friction is working against mg(sin theta) has anything to do with anything. If someone could point me in the right direction, I'd appreciate it.

 

[vincentchan]

friction force is the normal force times µ, on a horizontal surface, the normal force surely is equal to mg.. but your block is on an incline plane. you have to do a some calculation in order to find the normal..
and work done is Fd.. you knew that, right?

 

[Sny]

Normal force = mgcos (theta). (5kg)(9.8N)(cos 30) = 42.435

Force of friction = (42.435)(.436) = 18.502

Work = (Friction)(distance) = (18.502)(2.1) = 38.85

So, where'd I go wrong?

 

[dextercioby]

It looks alright...After all,the problem's asking about the work done by friction.Yyou could add a (-) sign to the final result,though,because of the negative acceleration due to friction.

Daniel.