Consider the system shown in Fig. P6.86. The rope and...

Consider the system shown in Fig. P6.86. The rope and Vpulley have negligible mass, and the pulley is frictionless. 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 $\mathrm{m} / \mathrm{s} .$ The blocks come to rest after moving 2.00 $\mathrm{m} .$ Use the work-energy theorem to calculate the coefficient of kinetic friction between the $8.00-\mathrm{kg}$ block and the tabletop.

Key Concepts

Energy Dissipation due to Friction

Energy dissipation due to friction refers to the conversion of mechanical energy (usually kinetic energy) into thermal energy as objects move against each other. This concept is important in applications of the work-energy theorem because it accounts for the non-conservative work done by friction, affecting the net energy available for motion.

Dynamics of Connected Bodies and Pulleys

The dynamics of connected bodies, such as those linked by a rope and pulley system, involve interdependent motion where the movement of one body directly affects the other. When pulleys are ideal (massless and frictionless) and ropes are inextensible, these constraints simplify the analysis, allowing the use of energy methods to relate the motions and forces acting on the individual bodies.

Work-Energy Theorem

The work-energy theorem states that the net work done on a system is equal to the change in its kinetic energy. This principle is widely used to analyze mechanical systems by linking forces acting on a body to its motion, making it especially useful when multiple forces, including non-conservative ones, are present.

Kinetic Friction

Kinetic friction is the force that opposes the relative motion between two surfaces in contact when they are sliding past each other. It is characterized by a coefficient (usually denoted µ?) which, when multiplied by the normal force between the surfaces, gives the magnitude of the frictional force. Understanding this concept is crucial for analyzing energy losses in systems involving motion.

Transcript

00:01 High in the given problem there are two blocks one of which has been put at the table top and it is having a mass m1 has 8 .00 kg another block is suspended with the help of thread which is passing over a frictionless pulley and its mass is m2 which is being given as 6 .00 kg its weight will be acting vertically downward creating a tension t in the string in the thread weight of this block is m1g so the normal reaction exerted by the tabletop on the block is n which is also equal to m1 g as the block will be moving towards right so force of frequency will be acting due left given by mu into n here mu is missing as the blocks are moving like this the suspended block is moving downward with an acceleration a so this block will also be moving towards right with the same acceleration a so if we use free body diagram of this second block having mass m2 we get to know m2 g might is the net force setting on it and using newton's second off motion it is put equal to m2 into a so the tension here comes out to be m2g minus m2a or we can say becomes m2g it becomes m2g minus a but before that we will have to find the acceleration in the system because to find this acceleration we will use kinematics because initial speed has been given to us as 0 .90 meter per second and the system is finally coming to rest so final speed is zero and the distance moved by the system by the hanging block and the block on the table is same which is given as 2 .00 meter.

02:40 So we use third equation of kinematics which says vf square is equal to v i square minus 2a s so it is converted into 0 is equal to 0 .9 to the whole square minus 2a into 2 so acceleration here comes out to be 0 .81 divided by by 4 what we can say acceleration is 0 .20 meter per second square so now going back to this equation we find tension t in the block in the in the in the thread as m2 into g minus a so this is m2 g minus a we know the mass of the block as 6 g 9 .8 minus 0 .2 so the tension comes out to be 57 .6 newton.

03:45 Now we use free body diagram of the block put on the table top.

03:51 It gives us t minus force of friction as m into a m1 into a and m1 is 8 so it becomes 8 a or we can say 8 into 0 .2 so tension we know 57 .6 for force of friction it becomes mean to n and n is 8g means 8 into 9 .8 is equal to 1 .6...

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