Two masses are connected by a rope over a frictionless...

Two masses are connected by a rope over a frictionless pulley. If the coefficient of kinetic friction between mass 1 (5.0 kg) and the table is 0.18, while mass 2 measures 3.2 kg, calculate: The constant acceleration of mass 2, from rest. The tension in the rope acting on mass 1. The minimum value of s in which the system will stay at rest. The final velocity of the mass when it reaches the pulley, which is located 24 cm [R] from its initial position.

Transcript

00:01 Okay, so we've got these two masses.

00:03 One's hanging over a frictionless pulley, and the other one has a coefficient of friction with the table of 0 .18.

00:13 Mass 1 is 5 kilograms, mass 2 is 3 .2 kilograms.

00:16 We need to figure out a few things about this situation.

00:19 The first one is the acceleration of the system.

00:23 Now, m1 and m2 are both going to experience the same acceleration because they are tied together.

00:30 And they will move together.

00:32 So let's start by analyzing the net force on mass number one.

00:37 So the net force on mass one given by ma is equal to the tension force, which is going to be pulling it this way, minus the friction force.

00:48 So our tension force is in the string, and the friction force will be mu times the normal force, which is equal to the weight.

00:56 So mu times m1g.

00:58 We can solve this for ten.

01:00 Tension, which we will then use in our next equation, the sum of the forces on m2.

01:07 The net force on m2 is going to be m2a, and that's going to be equal to the weight force down minus the tension force up.

01:17 Plugging this in for the tension, we now have m2a equals m2g minus m1a minus m1a minus mu m1g.

01:27 We can solve this for the acceleration by first adding that m1a.

01:32 A term to both sides, factoring out a on the left and g on the right, since both of these terms have a g, we can factor that out, and then dividing by this combined mass term, so the mass of both blocks.

01:47 Now we have an expression for the acceleration as such.

01:53 We can plug our values into that, because we know everything over here, g is 9 .8, m2 is 3 .2, mu is 0 .18, m1 is 5.

02:05 And we plug everything in and calculate, and we should get an acceleration of 2 .75 meters per second squared.

02:15 Now that we have the acceleration, we can go back to this equation up here, equation number one, and we can plug that in for a to solve for the tension force in the string.

02:28 And the tension force here is going to be the same as the tension force here.

02:31 So our tension will be 22 .6 newton...

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