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 11-18-2006, 04:59 PM #1 Member   Join Date: May 2006 Posts: 826 Can Friction do Positive WorK? I saw a question that asked to give an example of friction doing +ve Work, how is this possible and Give me an example if it is... Reply
 11-18-2006, 05:02 PM #2 Member   Join Date: May 2005 Location: somewhere Posts: 707 If I remember correcly, work is relative. It can change signs depending on the reference points, so I guess friction can do positive work. Reply
 11-18-2006, 07:25 PM #3 Junior Member   Join Date: Dec 2005 Posts: 165 Work is a scalar and doesn't have a direction... so all work should be positive, really. Reply
 11-18-2006, 09:23 PM #4 Member   Join Date: May 2006 Posts: 826 Yes work IS a scalar But it DOES have a sign, The fact dthat it doesnt have direction doesnt mena that it doesnt have a sign lol. Reply
 11-18-2006, 09:29 PM #5 Senior Member   Join Date: Nov 2004 Location: UCLA Posts: 7,734 Think about a small block resting on a larger block. If I move the large block forward, the static friction causes the small block to move in the same direction as the displacement, so the work is positive. I can't think of an example of kinetic friction doing positive work, though. Reply
 11-18-2006, 10:15 PM #6 Senior Member   Join Date: Feb 2006 Posts: 1,013 warblersrule, i think that is more inertia than friction... work of friction < = (coefficient of friction) * force of the normal... when there is no force acting on the smaller block, friction does no work... xylem: force of friction is in the opposite direction of the motion... so if you move a block to the right (+) friction does - work... if you move the block to the left (-) the friction does + work Reply
 11-19-2006, 01:32 AM #7 Junior Member   Join Date: Oct 2006 Posts: 77 Friction does negative work. Work=friction*r*costheta when you sum the forces, friction is a negative force because postitive forces are in direction of motion. theta is the angle between friction and r, your displacement. Since friction is in the direction opposite displacement, theta= 180 and work is thus negative. Therefore, friction always does negative work. Reply
 11-19-2006, 11:22 AM #8 Senior Member   Join Date: Feb 2006 Posts: 1,013 topeka... but if the force on the object that makes it move is in the negative direction (i.e. left), the force of friction is in the positive direction, and therefore does positive work... Reply
 11-19-2006, 01:09 PM #9 Junior Member   Join Date: Oct 2006 Posts: 77 I see what you're saying, but the work is still negative even though the friction is positive. Work=r*friction. Friction and r are always in opposite directions. So if r is in the negative direction, then friction is in the positive direction. So work is still neagive because a negative *postive is still negative. Another way of looking at it is with the equation Work=magnitude of friction*magnitude of r*cos180. In this equation, you only look at the MAGNITUDES of friction and r, which are always positive. In this case, cos 180 gives you the negative sign and Work done by friction is still negative. The cos 180 tells you that friction is always opposite the direction of displacement. Reply
 11-19-2006, 06:23 PM #10 Senior Member   Join Date: Feb 2006 Posts: 1,013 Oh, you're right. I'm stupid. sorry. Reply
 11-20-2006, 10:44 PM #11 Junior Member   Join Date: Nov 2006 Posts: 263 mmmmmhhhhmmmmmm Reply
 11-20-2006, 10:54 PM #12 Senior Member   Join Date: Apr 2006 Posts: 1,879 the first few replies that i read are stupid. the answer is no. Reply
 11-23-2006, 09:59 AM #13 Junior Member   Join Date: Jul 2006 Posts: 123 What about a moving car? The only force acting on the wheels is the static frictional force between them and the road...it's acting in the same direction as the displacement, and thus is doing positive work. Reply
 11-23-2006, 01:16 PM #14 Senior Member   Join Date: Oct 2005 Location: New York City, NY Posts: 1,017 no static force is working against displacement- that's why it's stuck there for a bit until you put enough force; if it went with displacement, the car would go super accelerated. Reply

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