AP Homework: Friction Problem - Find Minimum µk & Acceleration, Tension

[007banshee]

Ok this is my AP Homework just one question

The block of mass M1 is on a rough horizontal table. The string that attaches it to the block of mass M2 passes over a frictionless pulley of negligible mass. The coefficient of kinetic friction µk between M1 and the table is less than the coefficient of static friction M1

(a) In terms of M1 and M2 determine the minimum value of µk that will prevent the blocks from moving?The blocks are set in motion byy giving M2 a momentary push. In terms of M1,M2,µk and g determine each of the following..

(b)The magnitude of the acceleration of M1?(c) The tension in the string?Okay well i have no idea where to start because there are no given masses only thing known is gravity and nothing more so I am really confused does anyone have any ideas..

K i think this might help but I've figured that friction = tension since nothing is moving or at least that's what i think.

 

[007banshee]

And I don't really need an answer or anything just some help on how to start.

 

[kyle8921]

I would approach this problem by first understanding the concept of friction and its relation to tension and acceleration. Friction is the force that opposes motion between two surfaces in contact, and it is directly related to the coefficient of friction (µ). The coefficient of friction is a constant value that represents the roughness of the surfaces in contact and determines the amount of friction between them.

In this problem, we are given that the coefficient of kinetic friction (µk) is less than the coefficient of static friction (µs). This means that once the blocks start moving, the friction between them will be less than the maximum static friction, and the blocks will continue to move. The minimum value of µk that will prevent the blocks from moving is when µk is equal to µs, which is the point of maximum static friction. Therefore, the minimum value of µk is equal to the coefficient of static friction (µs).

Now, to determine the acceleration of M1, we can use Newton's second law of motion, which states that the net force applied to an object is equal to its mass multiplied by its acceleration (F=ma). In this case, the net force acting on M1 is the tension in the string (T) minus the friction force (µkN, where N is the normal force). So, we can write the equation as: T - µkN = M1a.

Since we know that the blocks are set in motion by giving M2 a momentary push, we can assume that the tension in the string (T) is equal to the weight of M2 (M2g). Therefore, the equation becomes: M2g - µkN = M1a.

To determine the tension in the string, we can use Newton's second law again. The net force acting on M2 is the weight of M2 (M2g) minus the tension in the string (T). So, we can write the equation as: M2g - T = M2a.

Now, we have two equations and two unknowns (a and T). We can solve these equations simultaneously to find the values of acceleration and tension. However, we also need to consider the constraint that the blocks will not move until the minimum value of µk is reached. So, we can substitute the value of µk (equal to µs) in the equations and solve for a and T.

In conclusion, the minimum