Two blocks sliding down a rough inclined plane

[cyberhat]

*****Homework Statement ****

Two blocks, A and B, are sliding down an inclined (20 degrees) plane. Block A is sliding in front of block B with both of them touching. The blocks slide from a distance of 6.50m from the bottom (along the inclined plane, not the surface the plane rests on).

Block A has a mass of 5.00kg with a kinetic friction coefficient of .150.
Block B has a mass of 10.00kg with a kinetic friction coefficient of .200.


Find the Acceleration of the blocks going down the plane.

What is the final velocity of Block A at the bottom?



----------------------------------------------

Attempt at solution consisted of starting with the FBD's of both blocks.

From this point here are my questions:

1) It doesn't seem like Block B is really putting any force on Block A going down the incline. Do I even need to consider it when finding my Net Force values in the x-direction? (I set the incline as the x-axis and perpendicular to it is the y-axis.

2) What would be the x-components of force?

 

[Mathoholic!]

Given that the blocks slide together down the plane, you may consider the sum of the following equations:

m_agsinθ-F_fa+F_b→a=F_ra

m_bgsinθ-F_fb-F_a→b=F_rb

→ m_bgsinθ-F_fb-F_a→b+m_agsinθ-F_fa+F_b→a=F_ra+F_rb

By Newton's third law, those forces cancel each other out since they have the same magnitude but opposite directions, and so:

(m_a+m_b)gsinθ-F_fb-F_fa=F_r(a+b)

Now the force of friction is given by the following equation (with μ being the coefficient of kinetic friction and N the force exerted by the plane on the block):

F_f=-μN

(Note that: N=mgcosθ)

Substitute and you get:

(m_a+m_b)gsinθ-(m_aμ_a+m_bμ_b)cosθ=F_r(a+b)

By Newton's second law, use F=ma and complete the problem using cinematics:

x(t)=x₀+v₀t+(1/2)at²