calculate the coeffcient uk of kinetic friction between the left block and the incline?

Question

Two blocks are connected by a thin inextensible string over a frictionless massless pulley as shown on the picture below
The acceleration of gravity is 9.8 m/s^2 :
M1=21 kg
M2=11 kg
v = 3.5 m/s
a = 3.4 m/s^2
ANGLE=23DEGREE
Given that the two blocks move at 3.5 m/s
under an acceleration of 3.4 m/s^2, calculate
the coefficient uk of kinetic friction between
the left block and the incline.

Answer 1

Look, what I understood without watching any picture, is that there are two blocks moving down from the inclinade, and that they are connected by a string. So here we go, first you need to draw the plane incline with the two blocks, the first block is the one with 11 kg, the second block is the one with 21 kg.

Block by block.

On the block of 11 kg :

weight = 11g ; g = 9.8 m / s^2

decomposing the weight we find two forces :

11g*sin(23) and 11g*cos(23)

11gsin(23) is the one that makes the block accelerate

11gcos(23) is the normal force, so the friction force on first block : 11gcos(23)*u

Friction force = 11gcos(23)u

u = coefficient of kinetic friction

And we have the tension of the string, so if you have drown the forces, applying the second Newton's law :

11gsin(23) - T - 11gcos(23)*u = 11a

a = acceleration

T = tension

For the block with mass 21 kg.

Practically the same, we have :

weight = 21g

Decomposing it :

21gsin(23)

21gcos(23) >>> normal force

force of friction : 21gcos(23)*u

Tension = T

Applying the second Newton's Law :

T + 21gsin(23) - 21gcos(23)*u = 21a

Here, the tension is in the same direction with : 21gsin(23), that's why you need to plus those forces

Now you have :

T + 21gsin(23) - 21gcos(23)*u = 21a.....(**)

11gsin(23) - T - 11gcos(23)*u = 11a .....(*)

(*) + (**) :

32gsin(23) - 32gcos(23)*u = 32a

a = 3.4 m/s^2

32g(sin23 - cos23u) = 32*3.4

0.35 = sin23 - cos23*u

u = 0.044

That's it

Hope that helped.