A cord passing over a pulley connects two masses, as shown,...

A cord passing over a pulley connects two masses, as shown, where m1 = 3.60 kg and m2 = 6.90 kg. The system accelerates with a magnitude of 0.94 m/s^2. The coefficient of kinetic friction between the masses and the incline is the same for both masses. Assume the pulley is frictionless, and the cord is massless. (Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.) (a) What is the coefficient of kinetic friction? (b) What is the tension (in N) in the cord?

Transcript

00:01 This is the diagram shown here there is a double inclined plane and each side is inclined with equal angle of 35 degree and it is a frictionless massless pulley means connects two masses m1 equal to 3 .6 kg with m2 equal to 6 .9 kg.

00:27 Right and as m2 is more than m1 then the pulley will rotate in the means clockwise direction like this it means the m2 will go down and m1 will go up because of difference in the mass right now we need to draw the free body diagram and when it rotates the acceleration of the system a is equal to 0 .94 meter per second square.

01:14 It means the downward acceleration of m2 equal to the upward acceleration of m1 is 0 .94 meter per second square.

01:25 Now we will draw the free body diagram of m1 and m2.

01:33 For m1, the first force acting downward is its own weight m1g.

01:43 Second force acting is the normal force, means from the inclined plane, which is n1.

01:59 And if we extend it backward, this angle is also 35.

02:04 So this becomes component of m1g in this direction will be m1g cost 35 and component of the weight m1g in this direction which is perpendicular to the other direction is m1g sine 35 right now as this block m1 moves up the force of also acts here in the backward direction we call it force of friction one right now again the tension acts in the string tension is always in the upward direction and as the pulley is massless and it has no friction the tension in the both parts of the string is same so tension here is also deep right and tension meets at the pulley if the pulley is massless and frictionless.

03:25 Similarly for m2 the first force acts is m2g and second is the normal reaction n2 and if we extend this n2 backward this angle is also 35 degree this is m2 g and as this angle is also 303.

03:56 This is m2 g and as this angle is also the component of m2g in this direction will be m2g cos 35 and in the perpendicular direction it will be m2g sine 35 right and when now this this block will move up this m2 will move down so the force of friction will act in the upward direction opposite to the direction of the motion we call it f2.

04:40 Now let is the common acceleration of the system and we write equation of motion, equation for m1.

04:54 Now m1 is moving up, so it means net upward force acting on m1 is m1a, is the acceleration.

05:07 This is a net upward force...

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