Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum Jan 4th 2016, 03:43 AM #1 Member   Join Date: Jan 2015 Posts: 96 Work-energy theorem problem Hi everyone, I have a problem that has me stumped and would appreciate some pointers as to where I am going wrong and maybe point me in the right direction for solving the problem. The problem is in essence to use the "Work-Energy Theorem" to find the co-efficient of kinetic friction in a pulley system. Problem - We have an 8.00 kg-block on flat horizontal tabletop attached via a rope and pulley to a hanging 6.00 kg-block. The rope and pulley have negligible mass and the pulley is friction-less. Initially the 6.00 kg-block is moving downward and the 8.00 kg-block is moving to the right, both with a speed of 0.900 m/s. The blocks come to rest after moving 2.00 meters. Use the Work-Energy Theorem to calculate the coefficient of Kinetic friction between the 8.00 kg-block and the tabletop. Basically I used the two main equations of the Work-Energy Theorem to try to solve this. I first calculated the Work used in moving each block using the difference in kinetic energy over the distance travelled that is ... W= K(2) - K(1) = 1/2 mv(2) sqrd - 1/2 mv(1) sqrd for each block, and since both blocks come to rest, each of the equations above reduce to just 1/2mv(1) sqrd for each block. I then took the difference in the values for the work each block expended to be the work expended by friction force. Now since (for a constant force) WORK also equals Force x distance, I equated the Work difference above to be equal to the work expended by the Kinetic Friction Force. And so Work difference = Work(Friction force)=Kinetic friction x distance moved. From my calculations I got W(8kg-block)=3.24 J, W(6kg-block)=2.43 J giving difference of 0.81 J as the Work of Friction force. Now since W=f x distance then f=w/distance =0.81/2.0 = 0.405 J I now have a value for Friction force (f) and I then used the relationship Friction = Coefficient of Friction x Normal force .. in this case 0.405=coefficient x mg (Normal force for 8kg-block = weight of block ie. 'mg') So we have coefficient = w/mg = 0.405/(8x9.8) = 0.405/78.8 = 0.02 . However the answer at the back of the book gives coefficient = 0.786. I have tried doing this in slightly different ways and the nearest I get to the correct answer is .. 0.75 (which if you notice is just the mass-ratio between the 2 blocks) ??? So where have I gone wrong ? can anyone help ?   Jan 6th 2016, 08:50 AM #2 Physics Team   Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 The work done by friction must equal the change in KE of the two blocks plus the change in PE of the 6 Kg block as it falls 2 meters under gravity: W_f = 1/2(m1+m2)v_i^2 + m_2 g h = (1/2) (6+8)Kg(0.9 m/s)^2 + (6 Kg)(9.8m/s^2)(2m) = 123.3 J That amount of work = force of friction times distance, so: F_f = 123.3J/2m = 61.6 N Can you take it from here?  Tags problem, theorem, workenergy ,

,

,

the rope and the pulley have neglible mass and the pulley is frictionless

Click on a term to search for related topics.
 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post jackthehat Kinematics and Dynamics 1 Mar 25th 2016 09:18 AM jackthehat Kinematics and Dynamics 6 Mar 6th 2016 12:40 PM chippysteve Energy and Work 3 Aug 18th 2013 01:57 PM chocolatelover Advanced Mechanics 1 Oct 14th 2008 10:59 AM chocolatelover Kinematics and Dynamics 1 Oct 14th 2008 10:56 AM