Physics Help Forum Work-energy theorem problem

 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?

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

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