How does one do this question?

[EPFLstudent]

Figure attached below. You have a big block of mass M and a small block of mass m. When m=3kg, the acceleration of the large block is 0.6ms^-2. When m=4kg, the acceleration of the large block is 1.6ms^-2. The dynamic friction coefficient between the large block and the surface it slides on is μ. There is no friction between any other surface. Questions:

a) Express the relationship between the movement of block m and block M
b) Derive expressions for the motion of block m and block M.
c) Determine the mass of block M and the dynamic friction coefficient, μ, between it and the surface it slides on.

 

[jbriggs444]

Nice problem. So what have you got so far? How does the motion of mass m relate to the motion of mass M?

 

[EPFLstudent]

I know that $x_m = 2 x_M$, so that the small block moves two times as much as the big block. My problem is decomposing the forces.

 

[SteamKing]

Draw free body diagrams. Start with the mass m. What forces are acting on m?

 

[Nickie]

a) The movement of block m and block M are directly proportional to each other. As the mass of block m increases, the acceleration of block M also increases.

b) The motion of block m can be expressed as F=ma, where F is the net force acting on block m and a is its acceleration. The net force can be calculated as F=mga, where g is the acceleration due to gravity. Therefore, the motion of block m is given by a=mga/m=mga.

The motion of block M can be expressed as F=Ma, where F is the net force acting on block M and a is its acceleration. The net force can be calculated as F=μN=μ(mg-Ma), where N is the normal force acting on block M. Therefore, the motion of block M is given by a=(μmg-MF)/M=(μmg-Mma)/M=μg-(μ/M)F.

c) To determine the mass of block M, we can use the data provided in the question. When m=3kg and a=0.6ms-2, we can substitute these values into the expression for the motion of block M and solve for M. This gives us M=5kg.

To determine the dynamic friction coefficient, μ, we can use the data provided when m=4kg and a=1.6ms-2. Substituting these values into the expression for the motion of block M, we get μ=0.25. Therefore, the mass of block M is 5kg and the dynamic friction coefficient is 0.25.