Two blocks with masses of 4kg and 8kg

[pezzang]

Two blocks with masses of 4kg and 8kg are connected by a string and slide down a 30 degree inclined plane. The coefficient of kinetic friction between 4kg block and the plane is 0.2; that between 8kg block and the palne is 0.35.
A) calculate the acceleration of each block.
b) calculate teh tension in the string
c) what happens if the positions of the blocks are reversed, so the 4kg block is above teh 8kg block?

picture

4kg ------- 8kg
the inclination goes up as it goes right so


8kg
------
---
4kg-

Like that.
I assumed that the acceleration of block a and b are teh same. can i assume that? and i soloved it using ma equals w times sin30 - uwcos30 - T
and Ma = mg sin20 - .25 times mg cos30 - T

and so on,. I got 2.21 as the accelration. can anybody confirm and tell me how they got it? and for the rest too?
Thanks.

 

[Chi Meson]

THe accelerations would be the same in the first scenario, but this is the end result, not the initial assumption.

THe coefficients of friction are different, so one will drag more than the other.

THe force of friciton is equal to "mu" times the perpendicular componant of the weight (since that will be the magnitude of the normal force)

Find the net force that is parallel to the plane for both objects as a system, then divide that force by the total mass.

I got 0.718 m/s^2 as the acceleration

 

[M98Ranger]

A) To calculate the acceleration of each block, we can use the equation F=ma. The forces acting on each block are the weight (mg), the normal force (N), and the force of friction (f). Since the blocks are connected by a string, the tension (T) in the string is the same for both blocks.

For the 4kg block:
F = ma
mg sin30 - μN = ma
(4kg)(9.8m/s^2)sin30 - (0.2)(4kg)(9.8m/s^2)cos30 = (4kg)a
a = 2.21 m/s^2

For the 8kg block:
F = ma
mg sin30 - μN = ma
(8kg)(9.8m/s^2)sin30 - (0.35)(8kg)(9.8m/s^2)cos30 = (8kg)a
a = 2.21 m/s^2

B) To calculate the tension in the string, we can use the equation ΣF = ma, where ΣF is the sum of all the forces acting on the block. For both blocks, the tension is the same.

For the 4kg block:
ΣF = ma
T - μN = ma
T = ma + μN
= (4kg)(2.21m/s^2) + (0.2)(4kg)(9.8m/s^2)cos30
= 8.84N + 6.86N
= 15.7N

For the 8kg block:
ΣF = ma
T - μN = ma
T = ma + μN
= (8kg)(2.21m/s^2) + (0.35)(8kg)(9.8m/s^2)cos30
= 17.68N + 10.78N
= 28.46N

C) If the positions of the blocks are reversed, the 8kg block will now be above the 4kg block. This will not change the acceleration of the blocks since it is determined by the forces acting on them, not their positions. However, the tension in the string will change.

For the 8kg block:
ΣF = ma
T - μN = ma
T = ma